Brilliant Summation Contest Season-1 (Part 1)

It is well-known that \(\displaystyle -\frac{\ln(1-x)}{1-x} = \sum_{n=1}^\infty H_nx^n\). So

\[\begin{align} \sum_{n=1}^\infty \frac{(-1)^nH_n}{2n+1} &= \int_0^1 \sum_{n=1}^\infty H_n(-x^2)^n \ dx \\ &= \int_0^1 \frac{-\ln(1+x^2)}{1+x^2} \ dx & (\text{Substitute } u = \tan^{-1} x) \\ &= \int_0^{\pi/4} -\ln(1+\tan^2 u) \ du \\ &= \int_0^{\pi/4} -\ln(\sec^2 u) \ du \\ &= \int_0^{\pi/4} 2 \ln \cos u \ du \\ &= G - \frac{\pi}{2} \ln 2 \end{align}\]

where G is Catalan's constant.

Solution to Special Problem : \[\underline{\text{Method 1}}\]

\[\text{S}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}\]

\[=\displaystyle\sum_{r=0}^{n}(-1)^r\dfrac{r!(n-r)!}{n!}\]

\[=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}\]

\[=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left[(-1)^r r!(n-r+1)! + (-1)^r (r+1)!(n-r)! \right]\]

\[=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left[(-1)^r r!(n-r+1)! - (-1)^{r+1} (r+1)!(n-r)! \right]\]

\[\color{blue}{=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left( T_{r} - T_{r+1} \right) }\tag{*}\]

where \(T_{r}= (-1)^r r!(n-r+1)!\)

Clearly, \((*)\) is a telescoping series. Evaluating it, we have,

\[\text{S}=(-1)^nT_{n+1}+T_{0}\]

\[=(-1)^n\dfrac{(n+1)!}{n!(n+2)} + \dfrac{(n+1)!}{n!(n+2)}\]

Since \(n\) is even,

\[\color{red}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}\]

\[\underline{\text{Method 2}}\]

Consider the following Lemma.

Lemma : \[\color{brown}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}\]

where \(k\) is an odd positive integer.

Proof : \[\text{S}=\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}\]

\[=\displaystyle \sum_{r=0}^k\dfrac{(-1)^{k-r}}{\dbinom{k}{k-r}}\]

\[\displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)} \]

\[= -\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}\]

\[\implies \text{S}=-\text{S}\]

\[\implies \text{S}=0\]

This completes the proof of the Lemma.

Now, since \(n\) is even, \(n+1\) is odd.

Using the Lemma, we have,

\[\color{blue}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}\]

Let,

\[\color{red}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}\]

Operating \((1)+(2)\),

\[\implies \text{J}+0=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n+1}{r}} +\dfrac{(-1)}{\dbinom{n+1}{n+1}}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{n-r}}{\dbinom{n+1}{n-r}} -1 \quad \displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}\]

\[=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n+1}{r+1}} -1\]

\[=\displaystyle \sum_{r=0}^n(-1)^r\left\{\dfrac{1}{\dbinom{n}{r}} + \dfrac{r+1}{n+1} \cdot \dfrac{1}{\dbinom{n}{r}}\right\} -1 \quad \displaystyle\color{red}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}\]

\[\implies \text{J} = \displaystyle\left(\dfrac{n+2}{n+1}\right) \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}+ \dfrac{1}{n+1} \sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}} -1\]

\[\implies \text{J}+1 = \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{1}{n+1} \underbrace{\color{#66f}{\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}}}_{\huge{= \ \text{G} \ (\text{let})}} \]

\[ \implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{G}}{n+1} \tag{3} \]

Now,

\[\text{G}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}\]

\[=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{n-r}\cdot (n-r)}{\dbinom{n}{n-r}}\]

\[=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot (n-r)}{\dbinom{n}{r}}\]

\[=\displaystyle n\sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n}{r}} - \sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot r}{\dbinom{n}{r}}\]

\[\implies \text{G}=n\cdot \text{J}-\text{G}\]

\[\implies\color{red}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}\]

From \((3)\ \text{&} \ (4)\),

\[\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}\]

\[\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square\]

Start by writing \(\displaystyle \dbinom nr ^{-1} = (n+1) \int_0^1 u^r (1-u)^{n-r} \, du \).

Multiply both sides by \( (-1)^r \) and adding over \(r\) gives:

\[ \begin{eqnarray} \sum_{r=0}^n\dfrac{(-1)^r}{ \binom{n}{r}} &=& (n+1) \int_0^1 \sum_{r=0}^n (-u)^r (1-u)^{n-r} \, du \\ &=& (n+1) \int_0^1 \left [(1-u)^{n+1} + (-1)^n u^{n+1} \right ] \, du \\ &=& \dfrac{n+1}{n+2} ( 1 + (-1)^n ) = \begin{cases} 0 , \qquad \quad n \text{ odd} \\ \dfrac{n+1}{2n+4} , \quad n \text{ even} \end{cases} \end{eqnarray} \]

Solution Of Problem 2 :

Proposition : \[ \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + \zeta(3) \]

Proof : It is easy to see that

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n} = \operatorname{Li}_{2}(x) + \dfrac{1}{2} \ln^2(1-x) \]

Dividing by \(x\) and integrating, we have,

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \int \dfrac{\operatorname{Li}_{2}(x)}{x}\mathrm{d}x + \dfrac{1}{2} \int \dfrac{\ln^2(1-x)}{x}\mathrm{d}x \]

\[ = \operatorname{Li}_{3}(x) + \dfrac{1}{2} [\ln^2(1-x) \ln x] + \int \dfrac{\ln x \ln (1-x)}{1-x}\mathrm{d}x \]

Let \(x=1-t\)

\[ \implies \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li}_{3}(x) + \dfrac{1}{2} [\ln^2(1-x) \ln x] - \int \dfrac{\ln (1-t) \ln t}{t} \mathrm{d}t \]

Now,

\[\int \dfrac{\ln (1-t) \ln t}{t} \mathrm{d}t = \ln t \operatorname{Li}_{2}(t) - \int \dfrac{ \operatorname{Li}_{2}(t) }{t}\mathrm{d}t \]

\[ = \ln t \operatorname{Li}_{2}(t) - \operatorname{Li}_{3}(t) \]

Substituting back, we have,

\[ \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + C\]

Putting \(x=0\), we get \(C=\zeta(3)\), thus,

\[ \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n^2} = \operatorname{Li_{3}} (x) - \operatorname{Li}_{3} (1-x) + \ln(1-x) \operatorname{Li}_{2}(1-x) + \dfrac{1}{2} \ln x \ln^2 (1-x) + \zeta(3) \quad \square \]

Now,

\[\displaystyle \text{S} = \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(3n+\frac{1}{2} \right) \right) = -2\ln 2 \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} + \sum_{n = 1}^{\infty} \sum_{k=1}^{3n} \dfrac{2}{2k-1} \]

Also,

\[\sum_{k=1}^{3n} \dfrac{1}{2k-1} = \sum_{k=1}^{6n} \dfrac{1}{k} - \sum_{k=1}^{3n}\dfrac{1}{2k}\]

\[ \sum_{k=1}^{3n} \dfrac{1}{2k-1} = H_{6n} - \dfrac{1}{2} H_{3n} \]

\[ \implies \text{S} = \zeta(2) \ln 2 + \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} [2 H_{6n} - H_{3n}] \]

Using cube roots and sixth roots of unity on the proposition, we have,

\[ \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} [2 H_{6n} - H_{3n}] = \dfrac{63}{2} \zeta(3) -14 \pi G \]

\[\implies \text{S} = \boxed{\zeta(2) \ln 2 + \dfrac{63}{2} \zeta(3) -14 \pi G}\]

Solution to Problem 13:

Just to see how nasty the last integral in this problem can get, here are the associated indefinite integrals!

Start with \[ \begin{array}{rcl} B(x) & = & \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}x^{n+1} \; = \; \int_0^x \left(\int_0^u \sum_{n=1}^\infty H_n^{(2)}v^n\,dv\right)\,\frac{du}{u} \\ & = & \displaystyle \int_0^x \left(\int_0^u \frac{\mathrm{Li}_2(v)}{1-v}\,dv\right)\,\frac{du}{u} \; = \; \int_0^x \left(\int_v^x \frac{\mathrm{Li}_2(v)}{u(1-v)}\,du\right)\,dv \\ & = & \displaystyle \int_0^x \ln\big(\tfrac{x}{v}\big) \frac{\mathrm{Li}_2(v)}{1-v}\,dv \end{array} \] Now (differentiate back to check): \[ \begin{array}{rcl} \displaystyle \int_0^x \frac{\mathrm{Li}_2(v)}{1-v}\,dv & = & \displaystyle -\tfrac13\pi^2 \ln(1 - x) + \ln^2(1 - x) \ln x + \ln(1 - x)\mathrm{Li}_2(x) + 2\mathrm{Li}_3(1 - x) - 2 \zeta(3) \\ \displaystyle \int_0^x \frac{\mathrm{Li}_2(v)}{1-v}\ln v\,dv & = & \displaystyle \tfrac{1}{45}\pi^4 - \tfrac16 \pi^2 \ln^2(1 - x) + \tfrac14 \ln^4(1 - x) + \tfrac12\ln^2(1 - x) \ln^2x \\ & & \displaystyle + (\ln^2(1 - x) - \ln(1 - x) \ln x + \ln^2x) \mathrm{Li}_2(x) - \tfrac12 \mathrm{Li}_2(x)^2 \\ & & \displaystyle + \ln^2\left(\frac{x}{1-x}\right) \mathrm{Li}_2\left(-\frac{x}{1-x}\right) + 2 \ln(1 - x) \mathrm{Li}_3(1 - x) - 2 \ln x \mathrm{Li}_3(x) \\ & & \displaystyle + 2 \ln(1 - x) \mathrm{Li}_3\left(-\frac{x}{1-x}\right) - 2 \ln x \mathrm{Li}_3\left(-\frac{x}{1 - x}\right) - 2 \mathrm{Li}_4(1 - x) \\ & & \displaystyle + 2 \mathrm{Li}_4(x) + 2 \mathrm{Li}_4\left(-\frac{x}{1 - x}\right) \end{array} \] and hence \[ B(\tfrac12) \; = \; \tfrac{1}{1440}\pi^4 - \tfrac{1}{24}\pi^2 \ln^22 + \tfrac{1}{24}\ln^42 + \tfrac14\zeta(3)\ln 2 \] Now consider \[ \begin{array}{rcl} \displaystyle A(x) & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2x^n \; =\; \sum_{n=1}^\infty \left(H_{n-1}^{(2)} + \tfrac{1}{n^2}\right)^2 x^n \\ & = & \displaystyle\sum_{n=1}^\infty \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2 x^{n+1} + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ & =& \displaystyle xA(x) + 2 B(x) + \mathrm{Li}_4(x) \end{array} \] so that \[ \begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} & = & \displaystyle A(\tfrac12) \; =\; 4B(\tfrac12) + 2\mathrm{Li}_4(\tfrac12) \\ & = & \displaystyle \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 \end{array} \] as required.

Solution to Problem 13:

Consider the function: \(\displaystyle f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n }^{ (2) } \right) }^{ 2 }{ x }^{ n } } \).

\[f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n-1 }^{ (2) }+\frac { 1 }{ { n }^{ 2 } } \right) }^{ 2 }{ x }^{ n } } \]

\[f\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n-1 }^{ (2) } \right) }^{ 2 }{ x }^{ n } } +{ \text{Li} }_{ 4 }(x)+2\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n-1 }^{ (2) }{ x }^{ n } }{ { n }^{ 2 } } } \]

On changing the summation index from \(n\) to \(n+1\), we get:

\[f\left( x \right) =xf\left( x \right) +{ \text{Li} }_{ 4 }(x)+2\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \quad \quad (...1)\]

We have to find \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \)

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } =\int _{ 0 }^{ x }{ \left[ \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ { \left( n+1 \right) } } } \right] dx } \quad \quad (...2)\]

Now we have to find \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ { \left( n+1 \right) } } } \)

I'll use the identity: \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ n } } =\int _{ 0 }^{ x }{ \frac { { \text{Li} }_{ 2 }\left( t \right) }{ t\left( 1-t \right) } dt } \).

Now, we differentiate w.r.t. x and get:

\[\sum _{ n=1 }^{ \infty }{ { H }_{ n }^{ (2) }{ x }^{ n-1 } } =\frac { { \text{Li} }_{ 2 }\left( x \right) }{ x\left( 1-x \right) } \]

We multiply both sides by x:

\[\sum _{ n=1 }^{ \infty }{ { H }_{ n }^{ (2) }{ x }^{ n } } =\frac { {\text{Li} }_{ 2 }\left( x \right) }{ \left( 1-x \right) } \]

Now, we integrate both sides w.r.t. \(x\) from 0 to x:

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ n+1 } } =2{ \text{Li} }_{ 3 }\left( 1-x \right) -2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } -{ \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } -\ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 }-2\zeta \left( 3 \right) \]

On dividing both sides by \(x\):

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n } }{ n+1 } } =\frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { {\text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \]

Now plugging this in eqn \(2\), we get:

\[\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } =\int _{ 0 }^{ x }{ \left[ \frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{ \text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { { \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \right] dx } \]

Now plugging this in eqn \(1\), we get:

\[f\left( x \right) =\frac { { \text{Li} }_{ 4 }\left( x \right) }{ 1-x } +\frac { 2 }{ 1-x } \int _{ 0 }^{ x }{ \left[ \frac { 2{ \text{Li} }_{ 3 }\left( 1-x \right) }{ x } -\frac { 2{\text{Li} }_{ 2 }\left( 1-x \right) \ln { \left( 1-x \right) } }{ x } -\frac { { \text{Li} }_{ 2 }\left( x \right) \ln { \left( 1-x \right) } }{ x } -\frac { \ln { \left( x \right) } { \left( \ln { \left( 1-x \right) } \right) }^{ 2 } }{ x } -\frac { 2\zeta \left( 3 \right) }{ x } \right] dx } \]

Now we take \(x=\frac { 1 }{ 2 } \) and compute the integral. Then we finally get:

\[\boxed{\displaystyle \sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} \; = \; \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2}\]

The final integral is left to the reader's exercise. I didn't post it as it would've made this solution too long.

Solution To Problem 12 :

Proposition : \[ f(a,z) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}\]

Proof : Note that,

\[ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} \]

where \( U_{n} (x) \) is the Chebyshev Polynomial of the second kind.

\[ \implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1} \]

Using Ramanujan Master Theorem, we have,

\[ f(a,z) = \dfrac{\pi}{\sin \pi z} U_{-s} (a) \]

\[ = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square \]

Now, using Gamma Triplication Formula,

\[\text{S} = \sum_{n=0}^{\infty} \dbinom{3n}{n} x^n = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{x^{\frac{1}{3}}(1 + x^2)}{(x^2 + 2ax + 1)(x^2 - 2ax + 1)} \mathrm{d}x \]

where \(a = \dfrac{3}{2} \sqrt{3x} \)

Using Partial Fraction and the Proposition, we have,

\[\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right] \]

After simplification, we have,

\[ \text{S} = \dfrac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}} \quad \square \]

Solution to Problem 12:

By the multiplication formula for the Gamma function, \[ (3n)! \; = \; \Gamma(3n+1) \; = \; \tfrac{1}{2\pi} 3^{2n+\frac12} \Gamma(n+\tfrac13)\Gamma(n+\tfrac23)\Gamma(n+1) \] and hence \[ {3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma(n+\frac13)\Gamma(n+\frac23)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi}B(n+\tfrac23,n+\tfrac13) \] Thus \[ \begin{array}{rcl}\displaystyle S \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4\sin^23\alpha\, u(1-u)}\,du \end{array} \] where \(0 < \alpha < \tfrac16\pi\) is such that \(\sin3\alpha = \tfrac12\sqrt{27x}\). The substitutions \(u = \sin^2\theta\) and then \(x = \tan\theta\) yield \[ \begin{array}{rcl} S & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^{\frac12\pi} \frac{\sin^{-\frac23}\theta \cos^{-\frac43}\theta \times 2\sin\theta \cos\theta}{1 - \sin^23\alpha\, \sin^22\theta}\,d\theta \; = \; \displaystyle \frac{\sqrt{3}}{\pi}\int_0^{\frac12\pi} \frac{\tan^{\frac13}\theta}{1 - \sin^23\alpha\, \sin^22\theta}\,d\theta \\ & = & \displaystyle\frac{\sqrt{3}}{\pi} \int_0^\infty \frac{x^{\frac13}(1 + x^2)}{(1 + x^2)^2 - 4\sin^23\alpha x^2}\,dx \; = \; \displaystyle\frac{\sqrt{3}}{\pi} \int_0^\infty \frac{x^{\frac13}(1 + x^2)}{(x^2 + 2x \sin3\alpha + 1)(x^2 - 2x \sin3\alpha + 1)}\,dx \\ & = & \displaystyle \frac{\sqrt{3}}{\pi} \int_0^\infty x^{\frac13}f(x)\,dx \; = \; \frac{\sqrt{3}}{\pi}I \end{array} \] where the meromorphic function \[ f(z) \; =\; \frac{z^2 + 1}{(z^2 + 2z \sin3\alpha + 1)(z^2 - 2z \sin3\alpha + 1)} \] has simple poles at each of the points in \(\mathcal{R} \,=\, \{ \pm ie^{3i\alpha}, \pm i e^{-3i\alpha}\}\).

Cutting the complex plane along the positive real axis, and integrating around the "keyhole contour" \(\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4\) where

• \(\gamma_1\) is the straight line segment from \(\epsilon\) to \(R\), just above the cut,
• \(\gamma_2\) is the circular segment \(z = Re^{i\theta}\) for \(0 \le \theta \le 2\pi\),
• \(\gamma_3\) is the straight line segment from \(\epsilon e^{2\pi i}\) to \(R e^{2 \pi i}\), just below the cut,
• \(\gamma_4\) is the circular segment \(z = \epsilon e^{i\theta}\) for \(0 \le \theta \le 2\pi\)

for any \(0 < \epsilon < 1 < R\), then \[ \left(\int_{\gamma_1} + \int_{\gamma_2} - \int_{\gamma_3} - \int_{\gamma_4}\right) z^{\frac13} f(z\,dz \; = \; 2\pi i \sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] so that \[ (1 - \omega)\int_\epsilon^R x^{\frac13}f(x)\,dx + \left(\int_{\gamma_2} - \int_{\gamma_4}\right) z^{\frac13}f(z)\,dz \; = \; 2\pi i\sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] Letting \(\epsilon \to 0\) and \(R \to \infty\), we obtain \[ (1 - \omega)I \; = \; 2\pi i\sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) \] and hence \[ I \; = \; \frac{\pi}{\sqrt{3}(2\cos2\alpha - 1)} \; = \; \frac{\pi}{\sqrt{3}(4\cos^2\alpha - 3)} \; = \; \frac{\pi \cos\alpha}{\sqrt{3}\cos3\alpha} \] so that \[ S \; = \; \frac{\cos\alpha}{\cos3\alpha} \; = \; \frac{2\cos\left(\frac13\sin^{-1}\left(\frac{3\sqrt{3x}}{2}\right)\right)}{\sqrt{4 - 27x}} \] as required.

Consider the binomial expansion:

\[\sum_{r=0}^n (-x)^{r} \dbinom{n}{r} =\left(1-x\right)^{n} \]

We'll write it as:

\[\sum_{r=1}^n (-x)^{r} \dbinom{n}{r} =\left(1-x\right)^{n}-1 \]

On dividing both the sides by \(-x\), we get:

\[\sum_{r=1}^n (-x)^{r-1} \dbinom{n}{r} =\frac{1-\left(1-x\right)^{n}}{x} \]

On integrating both the sides w.r.t. \(x\) from 0 to 1, we get:

\[\sum_{r=1}^n \frac{(-1)^{r-1}}r \dbinom{n}{r} =\int _{ 0 }^{ 1 }{ \frac { 1-{ \left( 1-x \right) }^{ n } }{ x } dx } =\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ n } }{ 1-x } dx } ={ H }_{ n }\\ \]

Solution To Problem 6 :

\[ \text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \dbinom{n}{r} \]

Note that, \( \displaystyle \sum_{k=r}^{n} \dbinom{k-1}{r-1} = \dbinom{n}{r} \), it follows,

\[\text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \sum_{k=r}^{n} \dbinom{k-1}{r-1} \]

Since \( \displaystyle \dbinom{k-1}{r-1} = \dfrac{r}{k} \dbinom{k}{r} \), we have,

\[ \text{S} = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r} \]

Interchanging order of summation, we have,

\[\text{S} = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r} \]

\[ = \sum_{k=1}^{n} \dfrac{1}{k} \]

\[ = H_{n} \quad \square \]

Call the given sum \(S\).

Then, by using the definition of trigonometric functions via complex exponentials, we have \[ S = \frac{1}{4i} \sum_{r=0}^{n} {n \choose r} \left( e^{inx}-e^{-inx} + e^{(2r-n)ix} - e^{-(2r-n)ix} \right) \]

Now, using binomial theorem and a bit of simplification, we get that \[ S=2^{n-1} \sin (nx) \]

PS: I don't have good summation problems. So, anyone can post in my behalf. :)

Solution To Problem 8 :

Lemma : \[\sum_{n=0}^{\infty} \dfrac{1}{n^2+1} = \dfrac{1}{2} (1 + \pi \coth(\pi)) \]

Proof : Let \[\text{S} = \sum_{n=0}^{\infty} \dfrac{1}{n^2+1}\]

\[\implies \text{S} = \dfrac{1}{2i} \sum_{n=0}^{\infty} \left(\dfrac{1}{n-i} - \dfrac{1}{n+i}\right) \]

Using Digamma Regularization, we have,

\[\text{S} = \dfrac{1}{2i} [\psi(i) - \psi(-i)] \]

Simplifying using Digamma Reflection Formula, we have,

\[\text{S} = \dfrac{1}{2} (1 + i \pi \cot(i\pi)) \]

\[ = \dfrac{1}{2} (1 + \pi \coth(\pi)) \quad \square \]

Now,

\[ \text{J} = \sum _{n=0}^{\infty} \dfrac{1}{n^4+4} \]

\[ = \sum _{n=0}^{\infty} \dfrac{1}{(n^2+2n+2)(n^2-2n+2)} \]

Using Partial Fraction,

\[ \text{J} = \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{(n^2-2n+2)} + \dfrac{1}{8} \sum _{n=0}^{\infty} \left( \dfrac{n+2}{n^2+2n +2} - \dfrac{n}{n^2-2n+2} \right) \]

\[ =\dfrac{1}{8} + \dfrac{1}{4} \sum _{n=1}^{\infty} \dfrac{1}{(n-1)^2+1} + \dfrac{1}{8} \sum _{n=0}^{\infty} \left( \dfrac{n+2}{n^2+2n +2} - \dfrac{n}{n^2-2n+2} \right) \]

\[ =\dfrac{1}{8} + \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{n^2+1} +\dfrac{1}{8} \sum _{n=0}^{\infty} \left( T_{n+2} - T_{n} \right) \]

where \(T_{n} = \dfrac{n}{n^2-2n+2} \)

Note that the latter sum telescopes. Evaluating it, we have,

\[\sum _{n=0}^{\infty} \left( T_{n+2} - T_{n} \right) = -1\]

\[ \implies \text{J} = \dfrac{1}{4} \sum _{n=0}^{\infty} \dfrac{1}{n^2+1} \]

\[ = \dfrac{1}{8} (1 +\pi \coth(\pi)) \quad \square \]

Claim: \( \displaystyle \sum_{a=1}^\infty \frac {8b^4}{a^4+4b^4} = b \pi \coth (b \pi) - 1 \)

Proof: Read Brian Chen's comment here.

The result follows.

Solution to Problem 8:

For any positive integer \(N>2\), let \(C_N\) be the positively oriented square contour with corners at \(\pm(N+\tfrac12) \pm i(N+\tfrac12)\). Let \(R\) be the set \(\{1+i,1-i,-1-i,-1+i\}\). Then, since \(\pi \cot\pi z\) is a meromorphic function, periodic of period \(1\), with a simple pole of residue \(1\) at each integer, we see that \[ \begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz & = & \displaystyle \sum_{n=-N}^N \mathrm{Res}_{z=n}\frac{\pi \cot\pi z}{z^4+4} + \sum_{u \in R}\mathrm{Res}_{z=u}\frac{\pi \cot\pi z}{z^4+4} \\ & = & \displaystyle -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} + \sum_{u \in R} \mathrm{Res}_{z=u} \frac{\pi \cot\pi z}{z^4 + 4} \\ & = & \displaystyle -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} - \tfrac{1}{16}\pi \sum_{u \in R} u \cot \pi u \end{array} \] Since \[ \cot \pi(\epsilon + \eta i) \; = \; - i \mathrm{coth}\,\pi\eta \] for \(\epsilon,\eta \in \{1,-1\}\), it follows that \[ \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz \; = \; -\tfrac14 + 2\sum_{n=0}^N \frac{1}{n^4 + 4} - \tfrac14\pi \mathrm{coth}\,\pi \] Since \(\pi \cot \pi z\) is uniformly bounded on \(C_N\) for all integers \(N\), we deduce that \[ \lim_{N \to \infty} \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{z^4 + 4}\,dz \; = \; 0 \] which gives the result.

\[S=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 }+4 } } \]

On splitting it using partial fractions, we get:

\[S=\frac { 1 }{ 4i } \left\{ \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-2i } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }+2i } } \right\} \]

We use the identity: \(\displaystyle\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}\). See the proof here.

On simplifying (exercise to readers), we get: \[\boxed{\displaystyle \sum_{n=0}^\infty \frac{1}{n^4 + 4} \; = \; \tfrac18(1 + \pi\mathrm{coth}\,\pi)} \]

Solution To Problem 16 :

Let \(a_{n} = \dfrac{1}{n}\) and \(b_{n} = \cos n \theta\). Note that both sequences satisfy the criterion of Dirchlet's Test. Therefore, the sum \( \displaystyle \text{S} = \sum_{n=1}^{\infty} \dfrac{\cos n \theta}{n}\) converges.

Now,

\[ \text{S} = \sum_{n=1}^{\infty} \dfrac{\cos n \theta}{n} \]

\[ = \Re \left( \sum_{n=1}^{\infty} \dfrac{e^{i n \theta}}{n} \right) \]

\[ = -\Re \left(\ln (1-e^{i \theta}) \right) \]

\[ = - \ln\left| 2\sin \left( \dfrac{\theta}{2} \right) \right| \quad \square \]

The contest has shifted to this note.

Solution to Problem 15:

Refer to my solution to Problem 8 for some of the technical details required about \(C_N\) and \(\pi \cot \pi z\).

Integrating \( \frac{\pi \cot \pi z}{(z^2+1)^2} \) around the positively-oriented square contour \(C_N\) with corners at \(\pm(N+\tfrac12) \pm i(N+\tfrac12)\), we deduce that \[\begin{array}{rcl} \displaystyle \frac{1}{2\pi i}\int_{C_N}\frac{\pi \cot \pi z}{(z^2+1)^2}\,dz & = & \displaystyle \left(\sum_{n =-N}^N\mathrm{Res}_{z=n} + \mathrm{Res}_{z=i} + \mathrm{Res}_{z=-i}\right) \frac{\pi \cot\pi z}{(z^2 + 1)^2}\,dz \\ & = & \displaystyle 1 + 2\sum_{n=1}^N \frac{1}{(n^2+1)^2} + \left(\mathrm{Res}_{z=i} + \mathrm{Res}_{z=-i}\right)\frac{\pi \cot\pi z}{(z^2 + 1)^2} \end{array} \] and \[ \mathrm{Res}_{z=\pm i}\frac{\pi \cot \pi z}{(z^2+1)^2} \; = \; \frac{d}{dz} \frac{\pi \cot \pi z}{(z \pm i)^2} \Big|_{z=\pm i} \; = \; -\tfrac14\pi^2 \mathrm{cosech}^2\pi - \tfrac14\pi\mathrm{coth}\pi \] so that \[ \frac{1}{2\pi i}\int_{C_N} \frac{\pi \cot \pi z}{(z^2+1)^2}\,dz \; = \; 1 + 2\sum_{n=1}^N \frac{1}{(n^2+1)^2} - \tfrac12\pi^2 \mathrm{cosech}^2\pi - \tfrac12\pi\mathrm{coth}\pi \] Since the integral around \(C_N\) tends to \(0\) as \(N \to \infty\), the result follows.

As before, I pass on setting the next problem.

Solution to Problem 14:

We note that \[ \begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{\psi^{(1)}(n)}{2^n n^2} & = & \displaystyle \tfrac12\zeta(2) - \sum_{n=1}^\infty \frac{1}{2^{n+1}(n+1)^2}\big[H^{(2)}_n - \zeta(2)\big] \\ & = & \displaystyle\zeta(2)\sum_{n=1}^\infty \frac{1}{2^n n^2} - \sum_{n=1}^\infty \frac{H_n^{(2)}}{2^{n+1}(n+1)^2} \\ & = & \displaystyle\zeta(2)\Big[ \tfrac{1}{12}\pi^2 - \tfrac12\ln^22\Big] - \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} \end{array} \] Going back to the previous problem, we have \[ \begin{array}{rcl} \displaystyle A(x) & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2x^n \; =\; \sum_{n=1}^\infty \left(H_{n-1}^{(2)} + \tfrac{1}{n^2}\right)^2 x^n \\ & = & \displaystyle \sum_{n=1}^\infty \big(H_n^{(2)}\big)^2 x^{n+1} + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ & =& \displaystyle xA(x) + 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \\ (1 - x)A(x) & = & \displaystyle 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} x^{n+1} + \mathrm{Li}_4(x) \end{array} \] and hence, putting \(x=\tfrac12\), \[ \tfrac12\sum_{n=1}^\infty \frac{\big(H_n^{(2)}\big)^2}{2^n} \; = \; 2\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} + \mathrm{Li}_4(\tfrac12) \] and so \[ \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2 2^{n+1}} \; = \; \tfrac{1}{1440}\pi^4 -\tfrac{1}{24}\pi^2\ln^22 + \tfrac{1}{24}\ln^42 + \tfrac14\zeta(3)\ln2 \] so that \[ \sum_{n=1}^\infty \frac{\psi^{(1)}(n)}{2^n n^2} \; = \; \tfrac{19}{1440}\pi^4 - \tfrac{1}{24}\pi^2\ln^22 - \tfrac{1}{24}\ln^42 - \tfrac14\zeta(3)\ln2 \] Someone else can post the next question.

Solution to Problem 10:

The sum is \[ S \; =\; \sum_{k \ge 1}\frac{\zeta(2k+1)-1}{2k+3} \; =\; \sum_{k \ge 1} \sum_{n \ge 2} \frac{1}{(2k+3)n^{2k+1}} \; = \; \sum_{n \ge 2} F\big(\tfrac{1}{n}\big) \] where \[ F(x) \; = \; \sum_{k \ge 1} \frac{x^{2k+1}}{2k+3} \; = \; \frac{1}{2x^2}\ln\left(\frac{1+x}{1-x}\right) - \tfrac13x - x^{-1} \qquad |x| < 1 \] Thus \[ S \; = \; \sum_{n \ge 2} \left(\tfrac12n^2\ln\left(\frac{n+1}{n-1}\right) - \frac{1}{3n} - n\right) \] After a shed-load of simplification, the partial sum \[ \begin{array}{rcl} S_N & = & \displaystyle \sum_{n = 2}^{N+1} \left(\tfrac12n^2\ln\big(\frac{n+1}{n-1}\big) - \frac{1}{3n} - n\right) \\ & = & \displaystyle \tfrac12(N+1)^2\ln(N+2) + \frac12N^2\ln(N+1) + 2\ln\left(\frac{G(N+1)}{\Gamma(N+1)^N}\right) \\ & & {} \displaystyle- \tfrac12\ln2 + \tfrac43 - \tfrac13H_{N+1} - \tfrac12(N+1)(N+2) \end{array} \] where \(G\) is the Barnes G-function. With the known asymptotics \[ \begin{array}{rcl} \displaystyle \ln G(N+1) & \sim & \displaystyle \tfrac{1}{12} - \ln A + \tfrac12N\ln(2\pi) + \tfrac12(N^2 - \tfrac16)\ln N - \tfrac34N^2 \\ \displaystyle \ln \Gamma(N+1) & \sim & (N+1)\ln(N+1) - (N+1) - \tfrac12\ln(N+1) + \tfrac12\ln(2\pi) + \frac{1}{12(N+1)} \end{array}\] as \(N \to \infty\), where \(A\) is the Glaisher constant, we deduce that \[ \ln\left(\frac{G(N+1)}{\Gamma(N+1)^N}\right) \; \sim \; -\ln A + \tfrac14N(N+4) + \tfrac12(N^2 - \tfrac16)\ln N - \tfrac12N(2N+1)\ln(N+1) \] as \(N \to \infty\) and hence we can deduce (after even more simplification) that \[ S \; = \; \lim_{N \to \infty}S_N \; =\; \tfrac{13}{12} - 2\ln A - \tfrac12\ln2 - \tfrac13\gamma \]

Solution To Problem 10 :

From the generating function of Riemann Zeta, we have,

\[ \sum_{n=1}^{\infty} \zeta(n+1) \ x^{n} = - \gamma - \psi(1-x) \]

\[ \implies \sum_{n=1}^{\infty} (1 + (-1)^{n}) \zeta(n+1) \ x^{n} = - 2 \gamma - \psi(1-x) - \psi(1+x) \]

\[ \implies \sum_{n=1}^{\infty} \zeta(2n+1) \ x^{2n+1} = -\gamma x - x \psi(x) - \dfrac{\pi x}{2} \cot (\pi x) - \dfrac{1}{2} \quad \quad (*) \]

Now,

\[ \text{S} = \sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{2k+3} \]

\[ = \int_{0}^{1} x \left[ \sum_{k=1}^{\infty} ( x^{2k+1} \zeta(2k+1) - x^{2k+1} ) \right] \mathrm{d}x \]

Using \((*)\), we have,

\[ \text{S} = - \int_{0}^{1} \left( \gamma x^2 + x^2 \psi (x) + \dfrac{x}{2} + \dfrac{\pi x^2}{2} \cot(\pi x) + \dfrac{x^4}{1-x^2} \right) \mathrm{d}x \]

\[ = \dfrac{13}{12} -2\log A - \dfrac{1}{2} \log 2 - \dfrac{\gamma}{3} \quad \square \]

Solution to Problem 7:

\[ \begin{array}{rcl} \displaystyle (1-z)\sum_{n=1}^\infty H_n^{(r)}z^n & = & \displaystyle \sum_{n=1}^\infty H^{(r)}_n z^n - \sum_{n=1}^\infty H^{(r)}_n z^{n+1} \\ & = & \displaystyle z + \sum_{n=2}^\infty \big[H^{(r)}_n - H^{(r)}_{n-1}\big]z^n \; = \; z + \sum_{n=2}^\infty \frac{z^n}{n^r} \\ & = & \displaystyle \mathrm{Li}_r(z) \end{array} \] for \(|z| < 1\).

Consider the following summation: \[ \begin{align} & (1-z) \sum_{n=1}^{\infty} H_n^{(r)} z^n \\ &= \sum_{n=1}^{\infty} (H_n^{(r)} - H_{n-1}^{(r)})z^n \quad (let \, H_0^{(r)}=0 )\\&= \text{Li}_r(z) \\ \large Q. E. D. \end{align} \]

Solution to Problem 5:

Since \[ F_n \; =\; \tfrac{1}{\sqrt{5}}\big[\phi^n - (-\phi^{-1})^n\big] \] we have \[ \begin{array}{rcl} \displaystyle \frac{1}{F_n F_{n-1}} & = & \displaystyle \frac{5}{\big(\phi^n - (-\phi^{-1})^n\big)\big(\phi^{n-1} - (-\phi^{-1})^{n-1}\big)} \; = \; \frac{5\phi^{2n-1}}{\big(\phi^{2n} - (-1)^n\big)\big(\phi^{2n-2} - (-1)^{n-1}\big)} \\ & = & \displaystyle \frac{5}{\phi + \phi^{-1}}\left[ \frac{1}{\phi^{2n} - (-1)^n} + \frac{1}{\phi^{2n-2} - (-1)^{n-1}}\right] \end{array} \] and hence (the series telescopes): \[ \begin{array}{rcl} \displaystyle \sum_{r=2}^\infty \frac{(-1)^n}{F_nF_{n-1}} & = & \displaystyle \frac{5}{\phi+\phi^{-1}}\sum_{n=2}^\infty \left[ \frac{(-1)^n}{\phi^{2n} - (-1)^n} - \frac{(-1)^{n-1}}{\phi^{2n-2} - (-1)^{n-1}}\right] \\ & = & \displaystyle \frac{5}{\phi + \phi^{-1}} \times \frac{1}{\phi^2 + 1} \; =\; \frac{5\phi}{(\phi^2 + 1)^2} \; = \; \frac{5\phi}{(\phi+2)^2} \\ & = & \displaystyle \frac{\phi}{\phi+1} \; = \; \phi - 1 \end{array} \] as required. I am happy to let Ishan set another problem; I am very busy at present.

Solution Of Problem 4 :

Let \[\text{S} = \sum_{\substack {0 \leq r \leq n \\ r \equiv 1 \pmod 3 }} {\dbinom{n}{r}} \]

Since,

\[\sum_{r=1}^{n} \dbinom{n}{r} x^r = (1+x)^n \]

\[ \implies \sum_{r=1}^{n} \dbinom{n}{r} {\omega}^r = (1+\omega)^n \tag{1} \]

\[\sum_{r=1}^{n} \dbinom{n}{r} {\omega}^{2r} = (1 + {\omega}^2)^n \tag{2} \]

\[ \sum_{r=1}^{n} \dbinom{n}{r} = 2^n \tag{3} \]

where \(\omega = e^{i \frac{\pi}{3}}\).

Operating \( {\omega}^2 \cdot (1) + \omega \cdot (2) + (3) \), we have,

\[\text{S} = \dfrac{1}{3} \left[ {\omega}^2(1+\omega)^n + \omega (1 + {\omega}^2)^n + 2^n \right] \]

Similarly, we get,

Generalization : \[ \sum_{\substack {0 \leq r \leq n \\ r \equiv k \pmod m }} {\dbinom{n}{r}} x^r = \sum_{r \geq 0} \dbinom{n}{rm + k} x^{rm+k} = \dfrac{1}{m} \sum_{r=0}^{m-1} {\omega}^{-rk} (1 + x\omega^r)^n \]

where \(\omega\) represents one of the non real \(m^{th}\) roots of unity.

Solution to Problem 9:

I'll use the generating function: \[\frac{\log^{2}(1-t)+Li_{2}(t)}{(1-t)}=\sum_{n=1}^{\infty}H_{n}^{2}t^{n}\]

Substitute: \(t=\frac{1}{2}\).

Hence the final answer is: \(\boxed{{ \left( \ln { 2 } \right) }^{ 2 }+\frac { { \pi }^{ 2 } }{ 6 } }\)