Brilliant Summation Contest Season-1 (Part 1)

Hi Brilliant! I've decided to start the first ever summation contest here.

The aims of the Summation contest are to improve skills in the computation of sums, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • You are NOT allowed to post a multiple summation problem.

  • Problems must be purely of summation. They shouldn't have integrals and products in them. Solutions can follow methods that use integrals and products.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • There is no restriction in the standard of summations.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

See Part-2.

Note by Aditya Kumar
3 years, 3 months ago

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Problem 2: Prove that n1(1)nn2(γ+ψ(3n+12))=632ζ(3)+ζ(2)log214πG\displaystyle \sum_{n\geq 1} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(3n+\frac12\right) \right)=\frac{63}{2}\zeta(3)+\zeta(2)\log 2 -14\pi G

GG is Catalan's constant.

This problem has been solved by Ishan Singh.

Proposition:

n1(1)nn2(γ+ψ(kn+12))=72k2ζ(3)+2πj=12k1jCl2(πj/k+π2k)πj=1k1jCl2(2πj/k+π/k)+ζ(2)log2\displaystyle \sum_{n\geq 1} \frac{(-1)^n}{n^2}\left(\gamma+\psi\left(kn+\frac12\right) \right)=\frac72k^2\zeta(3)+2\pi\sum_{j=1}^{2k-1}j\text{Cl}_2(\pi j/k+\frac{\pi}{2k}) - \pi\sum_{j=1}^{k-1}j\text{Cl}_2(2\pi j/k + \pi/k) + \zeta(2)\log 2

Aman Rajput - 3 years, 3 months ago

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Solution Of Problem 2 :

Proposition : n=1Hnxnn2=Li3(x)Li3(1x)+ln(1x)Li2(1x)+12lnxln2(1x)+ζ(3) \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

n=1Hnxnn=Li2(x)+12ln2(1x) \displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n} = \operatorname{Li}_{2}(x) + \dfrac{1}{2} \ln^2(1-x)

Dividing by xx and integrating, we have,

n=1Hnxnn2=Li2(x)xdx+12ln2(1x)xdx \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

=Li3(x)+12[ln2(1x)lnx]+lnxln(1x)1xdx = \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=1tx=1-t

    n=1Hnxnn2=Li3(x)+12[ln2(1x)lnx]ln(1t)lnttdt \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,

ln(1t)lnttdt=lntLi2(t)Li2(t)tdt\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

=lntLi2(t)Li3(t) = \ln t \operatorname{Li}_{2}(t) - \operatorname{Li}_{3}(t)

Substituting back, we have,

n=1Hnxnn2=Li3(x)Li3(1x)+ln(1x)Li2(1x)+12lnxln2(1x)+C \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=0x=0, we get C=ζ(3)C=\zeta(3), thus,

n=1Hnxnn2=Li3(x)Li3(1x)+ln(1x)Li2(1x)+12lnxln2(1x)+ζ(3) \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,

S=n=1(1)nn2(γ+ψ(3n+12))=2ln2n=1(1)nn2+n=1k=13n22k1\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,

k=13n12k1=k=16n1kk=13n12k\sum_{k=1}^{3n} \dfrac{1}{2k-1} = \sum_{k=1}^{6n} \dfrac{1}{k} - \sum_{k=1}^{3n}\dfrac{1}{2k}

k=13n12k1=H6n12H3n \sum_{k=1}^{3n} \dfrac{1}{2k-1} = H_{6n} - \dfrac{1}{2} H_{3n}

    S=ζ(2)ln2+n=1(1)nn2[2H6nH3n] \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,

n=1(1)nn2[2H6nH3n]=632ζ(3)14πG \sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2} [2 H_{6n} - H_{3n}] = \dfrac{63}{2} \zeta(3) -14 \pi G

    S=ζ(2)ln2+632ζ(3)14πG\implies \text{S} = \boxed{\zeta(2) \ln 2 + \dfrac{63}{2} \zeta(3) -14 \pi G}

Ishan Singh - 3 years, 3 months ago

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Special Problem :

Evaluate

r=0n(1)r(nr) \sum_{r=0}^{n} \dfrac{(-1)^r}{\dbinom{n}{r}}

Ishan Singh - 3 years, 3 months ago

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

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

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

=r=0n(1)rr!(nr)!(nr+1+r+1)n!×1(n+2)=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{#3D99F6}{(n-r+1+r+1)}}{n!}\color{#D61F06}{\times\dfrac{1}{(n+2)}}

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

=1n!(n+2)r=0n[(1)rr!(nr+1)!(1)r+1(r+1)!(nr)!]=\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]

=1n!(n+2)r=0n(TrTr+1)(*)\color{#3D99F6}{=\displaystyle \dfrac{1}{n!(n+2)} \sum_{r=0}^{n} \left( T_{r} - T_{r+1} \right) }\tag{*}

where Tr=(1)rr!(nr+1)!T_{r}= (-1)^r r!(n-r+1)!

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

S=(1)nTn+1+T0\text{S}=(-1)^nT_{n+1}+T_{0}

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

Since nn is even,

S=2n+1n+2\color{#D61F06}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}


Method 2\underline{\text{Method 2}}

Consider the following Lemma.

Lemma : r=0k(1)r(kr)=0\color{#624F41}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}

where kk is an odd positive integer.

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

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

(r=0nf(r)=r=0nf(nr))\displaystyle\color{#3D99F6}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)}

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

    S=S\implies \text{S}=-\text{S}

    S=0\implies \text{S}=0

This completes the proof of the Lemma.

Now, since nn is even, n+1n+1 is odd.

Using the Lemma, we have,

r=0n+1(1)r(n+1r)=0(1)\color{#3D99F6}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}

Let,

J=r=0n(1)r(nr)(2)\color{#D61F06}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}

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

    J+0=r=0n(1)r(nr)+r=0n+1(1)r(n+1r)\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}}

=r=0n(1)r(nr)+r=0n(1)r(n+1r)+(1)(n+1n+1)=\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}}

=r=0n(1)r(nr)+r=0n(1)nr(n+1nr)1(r=0nf(r)=r=0nf(nr))=\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{#3D99F6}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}

=r=0n(1)r(nr)+r=0n(1)r(n+1r+1)1=\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

=r=0n(1)r{1(nr)+r+1n+11(nr)}1[(nr)=nr(n1r1) & (nr)=(nnr)]=\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{#D61F06}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{\&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}

    J=(n+2n+1)r=0n(1)r(nr)+1n+1r=0n(1)rr(nr)1\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

    J+1=(n+2n+1)J+1n+1r=0n(1)rr(nr)= G (let)\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})}}

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

Now,

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

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

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

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

    G=nJG\implies \text{G}=n\cdot \text{J}-\text{G}

    G=Jn2(4)\implies\color{#D61F06}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}

From (3) & (4)(3)\ \text{\&} \ (4),

    J+1=(n+2n+1)J+Jn2(n+1)\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}

J=2n+1n+2\color{#20A900}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square

Ishan Singh - 3 years, 3 months ago

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Start by writing (nr)1=(n+1)01ur(1u)nrdu\displaystyle \dbinom nr ^{-1} = (n+1) \int_0^1 u^r (1-u)^{n-r} \, du .

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

r=0n(1)r(nr)=(n+1)01r=0n(u)r(1u)nrdu=(n+1)01[(1u)n+1+(1)nun+1]du=n+1n+2(1+(1)n)={0,n oddn+12n+4,n even \begin{aligned} \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{aligned}

Pi Han Goh - 3 years, 3 months ago

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This is the same method which Ishan talked about. Thanks for posting!

Aditya Kumar - 3 years, 3 months ago

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PROBLEM 11

Evaluate n=1(1)nHn2n+1  . \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1} \;.

This problem has been solved by Jake Lai and Aditya Kumar.

Mark Hennings - 3 years, 3 months ago

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

n=1(1)nHn2n+1=01n=1Hn(x2)n dx=01ln(1+x2)1+x2 dx(Substitute u=tan1x)=0π/4ln(1+tan2u) du=0π/4ln(sec2u) du=0π/42lncosu du=Gπ2ln2\begin{aligned} \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{aligned}

where G is Catalan's constant.

Jake Lai - 3 years, 3 months ago

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Reversing the order of integration and summation to obtain the second line of your argument takes a bit of justification. The generating function series is not monotonic, nor is it uniformly bounded by an integrable function, so the result is not automatic.

You need to integrate from 00 to x<1x < 1, obtaining N=1(1)nHn2n+1x2n+1  =  20tan1xlncosudu \sum_{N=1}^\infty \frac{(-1)^n H_n}{2n+1} x^{2n+1} \; =\; 2\int_0^{\tan^{-1}x} \ln \cos u\,du and then let xx tend to 11-, using Abel's Theorem to get the answer (having used the Alternating Series Test to prove that the series n(1)nHn2n+1\sum_n (-1)^n\frac{H_n}{2n+1} is convergent).

Anyway, you are up for the next question!

Mark Hennings - 3 years, 3 months ago

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Alternate Method:

n=1(1)n2n+1Hn=n=12H2n4n+1n=1Hn2n+1\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ 2n+1 } { H }_{ n } } =\sum _{ n=1 }^{ \infty }{ \frac { 2{ H }_{ 2n } }{ 4n+1 } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n } }{ 2n+1 } }

Then we can use appropriate generating functions for evaluating it. The integrals have to be carefully handled in order to get the solution.

Aditya Kumar - 3 years, 3 months ago

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SOLUTION OF PROBLEM 1 :

n0(1)nψ(3)(2n+2)=3!n0(1)nk01(k+2n+2)4=6n,k0(1)n(k+2n+2)4  =  6K0(n=0K/2(1)n)1(K+2)4=6K0,K0,1  (4)1(K+2)4=38K01(2K+1)4+6K01(4K+3)4=38(1116)190π4+3128ζ(4,32)=1256π4+1256ψ(3)(34) \begin{array}{rcl} \displaystyle \sum_{n \ge 0} (-1)^n \psi^{(3)}(2n+2) & = & \displaystyle 3!\sum_{n \ge 0} (-1)^n\sum_{k \ge 0}\frac{1}{(k+2n+2)^4} \\ & = & \displaystyle 6\sum_{n,k \ge 0} \frac{(-1)^n}{(k+2n+2)^4} \; = \; 6\sum_{K \ge 0} \left(\sum_{n=0}^{\lfloor K/2 \rfloor} (-1)^n\right) \frac{1}{(K+2)^4} \\ & = & \displaystyle 6\sum_{K \ge 0\,,\,K \equiv 0,1 \; (4)} \frac{1}{(K+2)^4} \\ & = & \displaystyle \tfrac38\sum_{K \ge 0} \frac{1}{(2K+1)^4} + 6\sum_{K \ge 0} \frac{1}{(4K+3)^4} \\ & = & \displaystyle \tfrac38\big(1 - \tfrac{1}{16}\big)\tfrac{1}{90}\pi^4 + \tfrac{3}{128}\zeta(4,\tfrac32) \\ & = & \tfrac{1}{256}\pi^4 + \tfrac{1}{256}\psi^{(3)}\big(\tfrac34\big) \end{array}

Mark Hennings - 3 years, 3 months ago

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Problem 3:
Evaluate :

r=0n[(nr)sinrxcos(nr)x]\sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right]

This problem has been solved by Deeparaj Bhat and Aditya Kumar.

Ishan Singh - 3 years, 3 months ago

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Call the given sum SS.

Then, by using the definition of trigonometric functions via complex exponentials, we have S=14ir=0n(nr)(einxeinx+e(2rn)ixe(2rn)ix) 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=2n1sin(nx) S=2^{n-1} \sin (nx)

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

Deeparaj Bhat - 3 years, 3 months ago

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Alternate Solution:

We use the identity: r=0nf(r)=r=0nf(nr)\displaystyle \sum _{ r=0 }^{ n }{ f\left( r \right) } =\sum _{ r=0 }^{ n }{ f\left( n-r \right) }

We take:

S=r=0n[(nr)sinrxcos(nr)x](1) S=\sum_{r=0}^n \left[\binom{n}{r}\cdot\sin rx \cdot \cos (n-r)x\right] \quad \quad (1)

On applying the identity, we get

S=r=0n[(nnr)sin(nrx)cosrx](2) S=\sum_{r=0}^n \left[\binom{n}{n-r}\cdot\sin (n-rx) \cdot \cos rx\right] \quad \quad (2)

On adding equations 1 and 2, we get:

2S=r=0n(nr)sinnx2S=\sum_{r=0}^n \binom{n}{r} \sin nx

Therefore, S=2n1sinnxS=2^{n-1} \sin nx

Aditya Kumar - 3 years, 3 months ago

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Problem 6 :

Prove That

r=1n(1)r1r(nr)=Hn\sum_{r=1}^n \frac{(-1)^{r-1}}r \dbinom{n}{r} =H_{n}

Notation : HnH_{n} denotes the Harmonic Number.

This problem has been solved by Aditya Kumar.

Ishan Singh - 3 years, 3 months ago

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Consider the binomial expansion:

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

We'll write it as:

r=1n(x)r(nr)=(1x)n1\sum_{r=1}^n (-x)^{r} \dbinom{n}{r} =\left(1-x\right)^{n}-1

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

r=1n(x)r1(nr)=1(1x)nx\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. xx from 0 to 1, we get:

r=1n(1)r1r(nr)=011(1x)nxdx=011xn1xdx=Hn\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 }\\

Aditya Kumar - 3 years, 3 months ago

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Solution To Problem 6 :

S=r=1n(1)r1r(nr) \text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \dbinom{n}{r}

Note that, k=rn(k1r1)=(nr) \displaystyle \sum_{k=r}^{n} \dbinom{k-1}{r-1} = \dbinom{n}{r} , it follows,

S=r=1n(1)r1rk=rn(k1r1)\text{S} = \sum_{r=1}^{n} \dfrac{(-1)^{r-1}}{r} \sum_{k=r}^{n} \dbinom{k-1}{r-1}

Since (k1r1)=rk(kr) \displaystyle \dbinom{k-1}{r-1} = \dfrac{r}{k} \dbinom{k}{r} , we have,

S=r=1nk=rn(1)r1k(kr) \text{S} = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r}

Interchanging order of summation, we have,

S=k=1nr=1k(1)r1k(kr)\text{S} = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{(-1)^{r-1}}{k} \dbinom{k}{r}

=k=1n1k = \sum_{k=1}^{n} \dfrac{1}{k}

=Hn = H_{n} \quad \square

Ishan Singh - 3 years, 3 months ago

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Problem 12:

Prove the following for x<427|x| < \dfrac{4}{27}.

n=0(3nn)xn=2cos(13sin1(33x2))427x\sum_{n=0}^\infty \binom{3n}{n}x^n = \frac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}}

This problem has been solved by Mark Hennings and Ishan Singh.

Jake Lai - 3 years, 3 months ago

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Solution to Problem 12:

By the multiplication formula for the Gamma function, (3n)!  =  Γ(3n+1)  =  12π32n+12Γ(n+13)Γ(n+23)Γ(n+1) (3n)! \; = \; \Gamma(3n+1) \; = \; \tfrac{1}{2\pi} 3^{2n+\frac12} \Gamma(n+\tfrac13)\Gamma(n+\tfrac23)\Gamma(n+1) and hence (3nn)  =  33n+122π×Γ(n+13)Γ(n+23)Γ(2n+1)  =  32n+122πB(n+23,n+13) {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 S  =  n=0(3nn)xn=32πn=032nxn01un13(1u)n23du=32π01u13(1u)23127xu(1u)du  =  32π01u13(1u)2314sin23αu(1u)du \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<α<16π0 < \alpha < \tfrac16\pi is such that sin3α=1227x\sin3\alpha = \tfrac12\sqrt{27x}. The substitutions u=sin2θu = \sin^2\theta and then x=tanθx = \tan\theta yield S=32π012πsin23θcos43θ×2sinθcosθ1sin23αsin22θdθ  =  3π012πtan13θ1sin23αsin22θdθ=3π0x13(1+x2)(1+x2)24sin23αx2dx  =  3π0x13(1+x2)(x2+2xsin3α+1)(x22xsin3α+1)dx=3π0x13f(x)dx  =  3πI \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)  =  z2+1(z2+2zsin3α+1)(z22zsin3α+1) 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 R={±ie3iα,±ie3iα}\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" γ1+γ2γ3γ4\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4 where

  • γ1\gamma_1 is the straight line segment from ϵ\epsilon to RR, just above the cut,
  • γ2\gamma_2 is the circular segment z=Reiθz = Re^{i\theta} for 0θ2π0 \le \theta \le 2\pi,
  • γ3\gamma_3 is the straight line segment from ϵe2πi\epsilon e^{2\pi i} to Re2πiR e^{2 \pi i}, just below the cut,
  • γ4\gamma_4 is the circular segment z=ϵeiθz = \epsilon e^{i\theta} for 0θ2π0 \le \theta \le 2\pi

for any 0<ϵ<1<R0 < \epsilon < 1 < R, then (γ1+γ2γ3γ4)z13f(zdz  =  2πiuRResz=uz13f(z) \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 (1ω)ϵRx13f(x)dx+(γ2γ4)z13f(z)dz  =  2πiuRResz=uz13f(z) (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 ϵ0\epsilon \to 0 and RR \to \infty, we obtain (1ω)I  =  2πiuRResz=uz13f(z) (1 - \omega)I \; = \; 2\pi i\sum_{u \in \mathcal{R}} \mathrm{Res}_{z=u} z^{\frac13}f(z) and hence I  =  π3(2cos2α1)  =  π3(4cos2α3)  =  πcosα3cos3α 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  =  cosαcos3α  =  2cos(13sin1(33x2))427x 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.

Mark Hennings - 3 years, 3 months ago

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Solution To Problem 12 :

Proposition : f(a,z)=0xzx2+2ax+1dx=πsinπzsin((1z)cos1(a))sin(cos1(a)) 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,

n=0Un(a)(x)n=1x2+2ax+1 \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1}

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

    n=0Un(a)Γ(n+1)(x)nn!=1x2+2ax+1 \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)=πsinπzUs(a) f(a,z) = \dfrac{\pi}{\sin \pi z} U_{-s} (a)

=πsinπzsin((1z)cos1(a))sin(cos1(a)) = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square

Now, using Gamma Triplication Formula,

S=n=0(3nn)xn=32π0x13(1+x2)(x2+2ax+1)(x22ax+1)dx\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=323xa = \dfrac{3}{2} \sqrt{3x}

Using Partial Fraction and the Proposition, we have,

S=32π[12f(13,a)+12f(73,a)14af(43,a)14af(103,a)+12f(13,a)+12f(73,a)+14af(43,a)+14af(103,a)]\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,

S=2cos(13sin1(33x2))427x \text{S} = \dfrac{2\cos(\frac{1}{3} \sin^{-1}(\frac{3\sqrt{3x}}{2}))}{\sqrt{4-27x}} \quad \square

Ishan Singh - 3 years, 3 months ago

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PROBLEM 13:

Show that n=1(Hn(2))22n  =  1360π416π2ln22+16ln42+2Li4(12)+ζ(3)ln2 \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

This problem has been solved by Aditya Kumar

Mark Hennings - 3 years, 3 months ago

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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 B(x)=n=1Hn(2)(n+1)2xn+1  =  0x(0un=1Hn(2)vndv)duu=0x(0uLi2(v)1vdv)duu  =  0x(vxLi2(v)u(1v)du)dv=0xln(xv)Li2(v)1vdv \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}