Brilliant Summation Contest Season-1 (Part 2)

Hi Brilliant! This is the sequel of the first part of Brilliant Summation Contest Season-1.

Update: This contest has been ended. Thanks for everyone's participation.

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.

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Note by Aditya Kumar
3 years, 3 months ago

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Problem 27:

Prove:

n=(eπ)n2=(π2)14Γ(34)\sum _{ n=-\infty }^{ \infty }{ { \left( -e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \frac { \pi }{ 2 } \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) }

Aditya Kumar - 3 years, 3 months ago

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We have a proof on brilliant

n=(eπ)n2=(π)14Γ(34)\sum _{ n=-\infty }^{ \infty }{ { \left( e^{-\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) }

Now use this formula ϕ(e2π/n)=(π)14Γ(34)n1/4r2,2/n221/4r2,2\phi(-e^{-2\pi/n})=\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \frac{n^{1/4}r_{2,2/n^2}}{2^{1/4}r_{2,2}}

rk,nr_{k,n} is a function of DedeKind Eta function. Which have some formulaes to solve.

r2,2=21/8andr2,1/2=21/8r_{2,2}=2^{1/8} \text{and} r_{2,1/2}=2^{-1/8}

Thus , you get (π2)1/4Γ(3/4)\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}

Aman Rajput - 3 years, 3 months ago

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At least could you elaborate on step 2?

Aditya Kumar - 3 years, 3 months ago

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@Aditya Kumar I used this identity from a ramanujan paper

Aman Rajput - 3 years, 3 months ago

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This is evaluated in ramanujan's notebook which I have but there's no proof for the formula used :/ . So I was trying to prove the above q-product

Aditya Narayan Sharma - 3 years, 3 months ago

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@Aditya Narayan Sharma share your q product .... if i have time i can solve it easily

Aman Rajput - 3 years, 3 months ago

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@Aman Rajput No, I m not talking bout the q-product actually. I m talking about the above formula , he used q-products and generalisations to come to this. So share if you have a proof of this , coz I didnt find

Aditya Narayan Sharma - 3 years, 3 months ago

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Problem 17 :

Prove That

n=02n1n!r=0n(4r+1)=π+2log(1+2)42 \displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}

This problem has been solved by Aditya Kumar.

Ishan Singh - 3 years, 3 months ago

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Solution to Problem 17:

S=n=02n1n!r=0n(4r+1)S=\sum _{ n=0 }^{ \infty } \frac { 2^{ n-1 }n! }{ \prod _{ r=0 }^{ n } (4r+1) }

S=n=0[2n1Γ(n+1)Γ(54)Γ(n+54)]S=\sum _{ n=0 }^{ \infty } \left[ \frac { { 2 }^{ -n-1 }\Gamma \left( n+1 \right) \Gamma \left( \frac { 5 }{ 4 } \right) }{ \Gamma \left( n+\frac { 5 }{ 4 } \right) } \right]

Using beta function, we get:

S=n=0[2n1B(n+1,54).(n+54)]S=\sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n-1 }B\left( n+1,\frac { 5 }{ 4 } \right) .\left( n+\frac { 5 }{ 4 } \right) \right]

S=12n=0[2nB(n+1,54).n]+58n=0[2nB(n+1,54)]S=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) .n \right] +\frac { 5 }{ 8 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ -n }B\left( n+1,\frac { 5 }{ 4 } \right) \right]

S=1201{n=0[2n.n.tn(1t)14]dt}+5801{n=0[2n.tn(1t)14]dt}S=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.n.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } +\frac { 5 }{ 8 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ -n }.{ t }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} }

Now I'll use the following geometric progression summations:

A=n=0[(xt)n(1t)14]=(1t)141xt,B=n=0[(xt)n.n.(1t)14]=xt(1t)14(1xt)2A=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ \left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 1-xt } ,\quad B=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ .n.\left( 1-t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { xt\left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 1-xt \right) }^{ 2 } }

S=5401(1t)142tdt+01x(1x)14(2x)2dxS=\frac { 5 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { { \left( 1-t \right) }^{ \frac { 1 }{ 4 } } }{ 2-t } dt } +\int _{ 0 }^{ 1 }{ \frac { x{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 2-x \right) }^{ 2 } } dx }

The first integral can be easily evaluated. The second one is also easily evaluated. I was screwed up and I couldn't evaluate it. So I asked for it here.

Hence the final answer on doing some simplification is:

n=02n1n!r=0n(4r+1)=π+2log(1+2)42\boxed{\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n-1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}}

Aditya Kumar - 3 years, 3 months ago

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Problem 18:

Prove that:

k=0n(n+knk)(1)n+k2k+1=22n+1cos(2(n1)π3)  ?\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}

This Problem has been solved by Ishan Singh.

Aditya Kumar - 3 years, 3 months ago

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Solution Of Problem 18 :

Proposition : Sn=r=0n(1)r(2n+1rr)=23sin(2(n+1)π3) \text{S}_{n} = \sum_{r=0}^{n} (-1)^r \dbinom{2n+1-r}{r} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)

Proof : Since (nr)=0 \dbinom{n}{r} = 0 for r>n r > n , we can rewrite the sum as

Sn=r=0(1)r(2n+1rr) \text{S}_{n} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r}

From the Binomial Theorem, we see that the sum is the coefficient of xnx^n in

f(x)=xn(1x)2n+1+xn1(1x)2n+f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots

=(1x)n+1x2x+1 = \dfrac{(1-x)^{n+1}}{x^2-x+1}

Note that,

n=0Ur(a)xr=1x22ax+1 \sum_{n=0}^{\infty}{{U}_{r}(a) {x}^{r}} = \dfrac{1}{x^2 -2ax+1}

where Ur(x) U_{r} (x) is the Chebyshev Polynomial of the second kind.

Putting a=12a = \dfrac{1}{2}, we see that,

f(x)=(1x)n+1n=0Ur(12)xr() f(x) = (1-x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*)

To calculate the coefficient of xnx^n in () (*) , we see that coefficient of xnrx^{n-r} for fixed rr in (1x)n+1(1-x)^{n+1} is (1)nr(n+1nr) (-1)^{n-r} \dbinom{n+1}{n-r} , so coefficient of xnx^n is,

r=0n(1)nr(n+1nr)Ur(12) \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n+1}{n-r} U_{r} \left(\dfrac{1}{2}\right)

Substituting rr1 r \mapsto r-1 , we have,

Sn=r=1n+1(1)n+1r(n+1n+1r)Ur1(12) \text{S}_{n} = \sum_{r=1}^{n+1} (-1)^{n+1-r} \dbinom{n+1}{n+1-r} U_{r-1} \left(\dfrac{1}{2}\right)

=(1)n+1r=1n+1(1)r(n+1r)Ur1(12)((nnr)=(nr)) = (-1)^{n+1} \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} U_{r-1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{n-r} = \dbinom{n}{r} \right)

=(1)n+1(23)r=1n+1(1)r(n+1r)sin(rπ3)(Ur1(12)=sin(rπ3)) = (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r-1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right)

(1)n+1(23)r=0n+1(1)r(n+1r)sin(rπ3)(sin0=0) (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right)

Now, r=0n+1(1)r(n+1r)sin(rπ3)=(r=0n+1(1)r(n+1r)eirπ3)=(1)n+1sin(2(n+1)π3)\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)

    Sn=23sin(2(n+1)π3) \implies \text{S}_{n} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square

Corollary : Sn=k=0n(1)n+k(n+1+k2k+1)=23sin(2(n+1)π3)\text{S}_{n} = \sum_{k=0}^n (-1)^{n+k} \dbinom{n+1+k}{2k+1} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right)

Proof : k=0n(1)n+k(n+1+k2k+1)=k=0n(1)n+k(n+1+knk)=k=0n(1)k(2n+1kk)=Sn\sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{2k+1} = \sum_{k=0}^n(-1)^{n+k} \dbinom{n+1+k}{n-k} = \sum_{k=0}^n(-1)^k \dbinom{2n+1-k}{k} = \text{S}_{n}

((nr)=(nnr) & r=0nf(k)=r=0nf(nk))\left( \because \dbinom{n}{r} = \dbinom{n}{n-r} \ \text{\&} \ \sum_{r=0}^{n} f(k) = \sum_{r=0}^{n} f(n-k) \right) \quad \square

Now,

λ=k=0n(1)n+k2k+1(n+knk) \lambda = \sum_{k=0}^n\dfrac{(-1)^{n+k}}{2k+1} \dbinom{n+k}{n-k}

=k=0n(1)n+k(n+k)!(2k+1)!(nk)! = \sum_{k=0}^n \dfrac{(-1)^{n+k}(n+k)!}{(2k+1)!(n-k)!}

=12n+1k=0n(1)n+k(n+k)![(n+k+1)+(nk)](2k+1)!(nk)! =\dfrac{1}{2n+1}\sum_{k=0}^n\dfrac{(-1)^{n+k}(n+k)!\left[(n+k+1)+(n-k)\right]}{(2k+1)!(n-k)!}

=12n+1k=0n((1)n+k(n+1+k2k+1)(1)n1+k(n+k2k+1)) =\dfrac{1}{2n+1}\sum_{k=0}^n\left((-1)^{n+k} \dbinom{n+1+k}{2k+1} - (-1)^{n-1+k} \dbinom{n+k}{2k+1} \right)

=12n+1(SnSn1) =\dfrac{1}{2n+1} \left(\text{S}_{n} - \text{S}_{n-1}\right)

Using the Proposition and simplifying, we have,

λ=22n+1cos(2(n1)π3) \lambda = -\dfrac{2}{2n+1} \, \cos\left(\frac{2(n-1)\pi}{3}\right) \quad \square


Note : Coefficient of xnx^n in the Proposition can also be found by splitting 1x2x+1\dfrac{1}{x^2-x+1} as partial fraction and expanding using Infinite GP.

Ishan Singh - 3 years, 3 months ago

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Problem 22:

Prove the identity:

k=1nHknk+1=(Hn+1)2Hn+1(2)\sum _{ k=1 }^{ n }{ \frac { { H }_{ k } }{ n-k+1 } } ={ \left( { H }_{ n+1 } \right) }^{ 2 }-{ H }_{ n+1 }^{ \left( 2 \right) }

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

Aditya Kumar - 3 years, 3 months ago

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Let S(r)=k=0nHrkk\displaystyle S(r)=\sum_{k=0}^{n} \frac{H_{r-k}}{k} for r>0r>0 with S(0)=0S(0)=0.

We have , S(n+1)=k=1nHknk+1=k=1nHnk+1k=k=1nHnkk+2Hnn+1=S(n)+2Hnn+1\displaystyle S(n+1) = \sum_{k=1}^{n} \frac{H_k}{n-k+1} = \sum_{k=1}^{n}\frac{H_{n-k+1}}{k} = \sum_{k=1}^{n}\frac{H_{n-k}}{k} + \frac{2H_n}{n+1} = S(n)+\frac{2H_n}{n+1}

Telescoping above we have , S(n+1)=k=0n2Hkk+1\displaystyle S(n+1) = \sum_{k=0}^{n}\frac{2H_k}{k+1}

Lemma :k=0n[(k+1)pkp]Hk(m)=(n+1)pHn(m)Hn(mp)\displaystyle \text{Lemma :} \large \boxed{\sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n}^{(m)}-H_{n}^{(m-p)}}

Proof :\text{Proof :} Using summation by parts we have ,

k=0n[(k+1)pkp]Hk(m)=(n+1)pHn+1(p)k=0n(k+1)(mp)=(n+1)pHn+1(p)Hn+1(mp)=(n+1)Hn(p)Hn(mp)\displaystyle \sum_{k=0}^{n} [(k+1)^p-k^p]H_{k}^{(m)} = (n+1)^p H_{n+1}^{(p)} - \sum_{k=0}^{n} (k+1)^{(m-p)} = (n+1)^p H_{n+1}^{(p)} - H_{n+1}^{(m-p)} = (n+1)H_{n}^{(p)} - H_{n}^{(m-p)} & thus proved.

We'll use a well known Harmonic sum identity , k=1nHkk=12((Hn)2+Hn(2))\displaystyle \sum_{k=1}^{n} \frac{H_k}{k} = \frac{1}{2}((H_n)^2+H_{n}^{(2)})

Applying the lemma for p=1,m=1p=-1,m=1 we have,

S(n+1)=2k=0nHkk+1=k=1n2Hkk+Hnn+1Hn(2)\displaystyle S(n+1) = 2\sum_{k=0}^{n} \frac{H_k}{k+1} = \sum_{k=1}^{n} \frac{2H_k}{k} + \frac{H_n}{n+1}-H_{n}^{(2)}

Simplifying we get , k=1nHknk+1=(Hn+1)2Hn+1(2)\displaystyle \sum_{k=1}^{n} \frac{H_k}{n-k+1} = (H_{n+1})^2-H_{n+1}^{(2)}

Aditya Narayan Sharma - 3 years, 3 months ago

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The generating function of the S(n)S(n) is n=1S(n)xn+1=n=1Hnxn×n=11nxn=ln2(1x)1x \sum_{n=1}^\infty S(n) x^{n+1} = \sum_{n=1}^\infty H_n x^n \times \sum_{n=1}^\infty \tfrac{1}{n} x^n = \frac{\ln^2(1-x)}{1-x} so the result follows using the known formulae for the generating functions for Hn(2)H_n^{(2)} and for (Hn)2(H_n)^2.

Mark Hennings - 3 years, 3 months ago

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We have,

S=k=1nHknk+1 \text{S} = \sum_{k=1}^{n}\dfrac{H_{k}}{n-k+1}

=k=1nr=1k1r(nk+1) = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{r(n-k+1)}

=r=1nk=rn1r(nk+1) = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(n-k+1)}

=r=1nk=rn1r(kr+1) = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(k-r+1)}

=r=1nk=rn(kr+1+r)r(kr+1)(k+1) = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(k-r+1 + r)}{r(k-r+1)(k+1)}

=r=1nk=rn1(k+1)[1(kr+1)+1r] = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right]

=k=1nr=1k1(k+1)[1(kr+1)+1r] = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{(k+1)} \left[\dfrac{1}{(k-r+1)} + \dfrac{1}{r}\right]

=2k=1nHkk+1 = 2\sum_{k=1}^{n} \dfrac{H_{k}}{k+1}

Substitute k+1k k+1 \mapsto k

    S=2k=2n+1Hk1k\implies \text{S} = 2\sum_{k=2}^{n+1} \dfrac{H_{k-1}}{k}

=2k=2n+1Hk11kk = 2\sum_{k=2}^{n+1} \dfrac{H_{k-1} - \dfrac{1}{k}}{k}

=2k=1n+1Hkk2Hn+1(2) = 2\sum_{k=1}^{n+1} \dfrac{H_{k}}{k} - 2H_{n+1}^{(2)}

Since k=1n+1Hkk=12(Hn+12+Hn+1(2))\displaystyle \sum_{k=1}^{n+1} \dfrac{H_{k}}{k} = \dfrac{1}{2} (H_{n+1}^2 + H_{n+1}^{(2)}) (I have proved it here), we have,

S=(Hn+1)2Hn+1(2)\text{S} = (H_{n+1})^2 - H_{n+1}^{(2)} \quad \square

Ishan Singh - 3 years, 3 months ago

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Problem 24:

Prove that k1F2kHk1(2)k2(2kk)=2π43755\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}

Fn Fibonacci number F_n - \text{ Fibonacci number }

This problem has been solved by Aditya Kumar.

Aman Rajput - 3 years, 3 months ago

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Solution of Problem 24:

Lemma 1: x2=12n=1[22nn2(2nn)sin2nx]{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }

Proof:

Consider: fn(x)=sin2nx(2n)!{ f }_{ n }\left( x \right) =\frac { \sin ^{ 2n }{ x } }{ \left( 2n \right) ! } .

Then we get: f"n(x)=fn1(x)(2n)2fn(x)\\ { f" }_{ n }\left( x \right) ={ f }_{ n-1 }\left( x \right) -{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right)

We can write x2=n=0anfn(x)\displaystyle { x }^{ 2 }=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ f }_{ n }\left( x \right) } for some function ana_n.

On differentiating both sides twice:

2=n=1an(fn1(x)(2n)2fn(x))=n=0an+1fn(x)n=1an(2n)2fn(x)2=\sum _{ n=1 }^{ \infty }{ { a }_{ n }\left( { f }_{ n-1 }\left( x \right) -{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) \right) } =\sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ f }_{ n }\left( x \right) } -\sum _{ n=1 }^{ \infty }{ { a }_{ n }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) }

From this relation a1=2a_1=2 and an+1=an(2n)2a_{n+1}=a_n{(2n)}^{2}

Thus, an2=k=1n1ak+1ak=22n2(n1)!2\displaystyle \frac { { a }_{ n } }{ 2 } =\prod _{ k=1 }^{ n-1 }{ \frac { { a }_{ k+1 } }{ { a }_{ k } } } ={ 2 }^{ 2n-2 }{ \left( n-1 \right) ! }^{ 2 }.

Hence, we get : x2=12n=1[22nn2(2nn)sin2nx]\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }

Lemma 2: x4=12n=1[22nsin2nxHn1(2)n2(2nn)]{ x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }

Proof:

From Lemma 1, we can write: x2=12n=1[22nn2(2nn)sin2nx]=12n=1bn2fn(x)\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) }

Here, bn=2n(n1)!b_n={ 2 }^{ n }(n-1)! and bn+1=2nbnb_{n+1}=2nb_n.

We can write: x4=12n=1anbn2fn(x)\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) } for some function ana_n.

On differentiating both sides twice:

12x2=12n=0an+1bn+12fn(x)12n=1anbn2(2n)2fn(x)12{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ b }_{ n+1 }^{ 2 }{ f }_{ n }\left( x \right) } -\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) }

Thus, we get: (an+1an)bn2(2n)2=12bn2({ a }_{ n+1 }-{ a }_{ n }){ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }=12{ b }_{ n }^{ 2 }

From here we get: (an+1an)=12(2n)2({ a }_{ n+1 }-{ a }_{ n })=\frac { 12 }{ { \left( 2n \right) }^{ 2 } } .

Hence, an=Hn1(2){ a }_{ n }={ H }_{ n-1 }^{ (2) }.

Therefore, we get: x4=12n=1[22nsin2nxHn1(2)n2(2nn)]\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n-1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }.

From lemma 2, we get: (sin1x)4=32n=122nHn1(2)n2(2nn)x2n ...(A)\displaystyle (\sin^{-1} x)^4 = \frac32 \sum_{n=1}^\infty \frac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2n} \ \quad ...(A)

Now, in the problem, we use: F2k=φ2k1φ2k5{ F }_{ 2k }=\frac { { \varphi }^{ 2k }-\frac { 1 }{ { \varphi }^{ 2k } } }{ \sqrt { 5 } }

On substituting and using AA, we get: k1F2kHk1(2)k2(2kk)=2π43755\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k-1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}

Aditya Kumar - 3 years, 3 months ago

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(+1) Nicely explained! I solved using Beta Functions and Integration.

Ishan Singh - 3 years, 3 months ago

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Problem 20 :

Evaluate im=1nim1=1imi2=1i3i1=1i21\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1

This problem has been solved by Aditya Sharma and Ameya Daigavane.

Deeparaj Bhat - 3 years, 3 months ago

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Although @Aditya Sharma has solved the question - I thought of an approach that might be considered more "illuminating" than induction.

Let's think of the i1,i2,...im i_1, i_2,...i_m as numbers on a register - each ik i_k is one number from 11 to nn. Initially, the register reads m m ones, because all the iki_k are 11.
For each unique number on the register, we add one to our count.

Now, we turn the first number (imi_m) to 22.
The register reads 22 followed by m1m - 1 11s.
Note that a digit to the right of another digit can never be greater than it, because ijiki_j \geq i_kif jkj \geq k.

So, the number on our register is an mm digit number whose digits are non-decreasing when read from left to right.
We can keep turning the register and counting how many unique numbers we get, until we reach mm digits with value nn. After this, our register stops working.

It is easy to see that every single number satisfying the non-decreasing digit condition is encountered, and only once.

So our question now becomes:
What is the number of tuples i1,i2,...im i_1, i_2,...i_m satisfying,

1i1i2imn 1 \leq i_1 \leq i_2 \leq \ldots \leq i_m \leq n

Because for each one of these tuples, we have a number on our register.

If we write the numbers in a row, i1i2im i_1 i_2 \ldots i_m and divide them into nn blocks, such that the numbers enclosed in the nthn^{th} block get value n n , then we have found one such possibility. We need n1n - 1 bars to separate the numbers into nn blocks.
For example, if m=4,n=4m = 4, n = 4, we can divide the numbers as, i1i2i3i4 i_1 | | i_2 i_3 | i_4 Here 0 i1=1,i2=i3=3,i4=4 i_1 = 1, i_2 = i_3 = 3, i_4 = 4

Clearly, the number of ways we can separate these numbers, is the number of tuples we need, because numbers with higher indices are always given higher (or equal) values.

So, we can finish with stars-and-bars, as we have mm numbers, and n1n -1 separators, so the total number of separations/tuples is: (m+n1n1) \binom {m + n - 1}{n - 1} which is the same answer as above.

Ameya Daigavane - 3 years, 3 months ago

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Let S=im=1nim1=1imi2=1i3i1=1i21S =\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1

S=im=1nim1=1imi2=1i3(i21)\displaystyle S=\sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \binom{i_2}{1}

S=im=1nim1=1imi3=1i4(i3+12)\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_3=1}^{i_4} \binom{i_3+1}{2}

S=im=1nim1=1imi4=1i5(i4+23)\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m-1}=1}^{i_m} \cdots \sum_{i_4=1}^{i_5} \binom{i_4+2}{3}

It's obvious from induction that S=im=1n(im+m2m1)=nm(m+n1n)\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m-2}{m-1} = \frac{n}{m}\binom{m+n-1}{n}

Aditya Narayan Sharma - 3 years, 3 months ago

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Problem 21:

Prove that: n=0(2nn)5n=5\large \displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}

This problem has been solved by Aditya Kumar.

Aditya Narayan Sharma - 3 years, 3 months ago

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Solution to Problem 21:

We use the well known generating function of Central Binomial coefficient:

114x=n=0(2n)!(n!)2xn\frac{1}{\sqrt{1 - 4x}} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n

Here substitute x=15x=\frac{1}{5}.

Hence, n=0(2nn)5n=5\large \boxed{\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{-n} = \sqrt{5}}

Aditya Kumar - 3 years, 3 months ago

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Proof of that generating function?Thanks.

Harsh Shrivastava - 3 years, 3 months ago

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@Harsh Shrivastava There are many ways but I learnt it here.

Aditya Kumar - 3 years, 3 months ago

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Problem 29:

Prove:

m=199sin(17mπ100)sin(39mπ100)1+cos(mπ100)=1037\displaystyle \sum\limits_{m=1}^{99}{\frac{\sin{\left(\frac{17 m \pi}{100}\right)} \sin{\left(\frac{39 m \pi}{100}\right)}}{1+\cos{\left( \frac{m\pi}{100} \right) }}}=1037

Due to time constraint, I have decided to post the solution.

Jack Lam - 3 years, 3 months ago

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Define T(k)=m=1n1cosmkπn1+cosmπn \displaystyle T(k) = \sum_{m=1}^{n-1} \frac{\cos{\frac{mk\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}

Observing that cos(k+1)θ+2coskθ+cos(k1)θ=2coskθ(1+cosθ)\cos{(k+1)\theta} + 2\cos{k\theta} + \cos{(k-1)\theta} = 2\cos{k\theta}(1+\cos{\theta})

The following recurrence relation is obtained:

T(k+1)+2T(k)+T(k1)=2m=1n1cosmkπn=1+coskπ+sinkπtankπ2n\displaystyle T(k+1) + 2T(k) + T(k-1) = 2 \sum_{m=1}^{n-1} \cos{\frac{mk\pi}{n}} = 1+\cos{k\pi} + \sin{k\pi} \tan{\frac{k\pi}{2n}}

Note that an implicit restriction of knZ k \notin n \mathbb{Z} is forced because the GP formula is invalid when the common ratio is ±1\pm 1.

Evaluate the recurrence relation at k+1k+1 and add together to yield

T(k+2)+3T(k+1)+3T(k)+T(k1)=2 T(k+2) + 3T(k+1) + 3T(k) + T(k-1) = 2

Evaluate this recurrence relation at kk and take the difference to homogenise the recurrence relation:

T(k+2)+2T(k+1)2T(k1)T(k2)=0 T(k+2) + 2T(k+1) - 2T(k-1) - T(k-2) = 0

The characteristic roots are 11 and 1-1 with multiplicity 33.

The general solution to the recurrence relation is thus

(C1k2+C2k+C3)(1)k+C4 (C_1 k^2 + C_2 k+C_3)(-1)^k + C_4

Returning to the original problem, define S(a,b,n)=m=1n1sinamπnsinbmπn1+cosmπn\displaystyle S(a,b,n) = \sum_{m=1}^{n-1} \frac{\sin{\frac{am\pi}{n}}\sin{\frac{bm\pi}{n}}}{1+\cos{\frac{m\pi}{n}}} where 0ban 0 \le b \le a \le n

We find that 2S(a,b,n)=T(ab)T(a+b)2S(a,b,n) = T(a-b)-T(a+b)

Note that (1)a+b(1)ab (-1)^{a+b} \equiv (-1)^{a-b}

Substituting T(k)T(k) into the above and expanding, we obtain

S(a,b,n)=(2C1a+C2)b(1)a+b+1=(1)a+bC1^b(C2^a)\displaystyle S(a,b,n) = (2C_1 a + C_2)b(-1)^{a+b+1} = (-1)^{a+b}\hat{C_1}b(\hat{C_2}-a)

A short computation yields S(1,1,n)=n1S(1,1,n) = n-1

From S(n,b,n)=0&S(1,1,n)>0 S(n,b,n) = 0 \, \& \, S(1,1,n)>0, C2^=n\hat{C_2} = n

From S(1,1,n)=n1S(1,1,n) = n-1, C1^=1\hat{C_1} = 1

Therefore, we conclude S(a,b,n)=(1)a+bb(na) S(a,b,n) = (-1)^{a+b} b(n-a)

Substituting b=17,a=39,n=100b=17, a=39, n=100, and the result which was to be shown follows.

Jack Lam - 3 years, 3 months ago

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Nice solution. I missed the 48 hour time.

Aditya Kumar - 3 years, 3 months ago

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This was the best solution!

Deeparaj Bhat - 3 years, 3 months ago

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Problem 23 :\text{Problem 23 :}

Prove that : n=(1)n(n+a)4=π46cos(πa)sin2(πa)(6csc2(πa)1)\displaystyle \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(n+a)^4} = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) -1) for aRa \in \mathbb{R} and aa is not an integer.

This problem has been solved by Aman Rajput.

Aditya Narayan Sharma - 3 years, 3 months ago

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n(1)n(n+a)4=(1)442(k01(k+1a2)4+1(k+a2)41(k1a2)41(k1+a2)4)\displaystyle \sum_{n\geq -\infty} \frac{(-1)^n}{(n+a)^4}=\frac{(-1)^4}{4^2}\left( \sum_{k \geq 0}\frac{1}{(k+1-\frac{a}2)^4}+\frac{1}{(k+\frac{a}2)^4}-\frac{1}{(k-\frac{1-a}2)^4}-\frac{1}{(k-\frac{1+a}2)^4} \right)

=116(16(ψ(3)(1a2)+ψ(3)(a2)ψ(3)(1a2)ψ(3)(1+a2)))\frac{1}{16} (\frac16(\psi^{(3)}(1-\frac{a}2)+\psi^{(3)}(\frac{a}2)-\psi^{(3)}(\frac{1-a}2)-\psi^{(3)}(\frac{1+a}2)))

Using reflection formula and On a little bit solving you get

π46cos(πa)sin2(πa)(6csc2(πa)1)\frac{\pi^4}{6}\frac{cos(\pi a)}{\sin^2(\pi a)}(6\csc^2(\pi a)-1) Sorry for short solution

Aman Rajput - 3 years, 3 months ago

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Yup its ok. Nice solution .

Aditya Narayan Sharma - 3 years, 3 months ago

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Alternative :\text{Alternative :}

Consider the function f(z)=1(z+a)4\displaystyle f(z)=\frac{1}{(z+a)^4} which analytic and has poles at the point z=az=-a of order 4.

Considering the square contour positively oriented having vertices ±(1+i) \pm (1+i) and by residue lemma we derive ,

Res(a)=16limzad3dz3(πcsc(πz)) \displaystyle Res(-a) = \frac{1}{6} \lim_{z\to -a} \frac{d^3}{dz^3}(π\csc(πz))