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, 6 months ago

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PROBLEM 16: If 0<θ<2π0 < \theta < 2\pi, evaluate n=1cosnθn \sum_{n=1}^\infty \frac{\cos n\theta}{n} Don't forget to justify the convergence!

Mark Hennings - 3 years, 6 months ago

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The contest has shifted to this note.

Aditya Kumar - 3 years, 6 months ago

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Solution To Problem 16 :

Let an=1na_{n} = \dfrac{1}{n} and bn=cosnθb_{n} = \cos n \theta. Note that both sequences satisfy the criterion of Dirchlet's Test. Therefore, the sum S=n=1cosnθn \displaystyle \text{S} = \sum_{n=1}^{\infty} \dfrac{\cos n \theta}{n} converges.

Now,

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

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

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

=ln2sin(θ2) = - \ln\left| 2\sin \left( \dfrac{\theta}{2} \right) \right| \quad \square

Ishan Singh - 3 years, 6 months ago

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The next problem would be posted here.

Aditya Kumar - 3 years, 6 months ago

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Problem 15:

Prove that:

n=11(n2+1)2=π4(coth(π)+πcsch2(π)2π)\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n^{2}+1)^{2}}=\frac{\pi}{4}\left(\coth(\pi)+\pi \text{csch}^{2}(\pi)-\frac{2}{\pi}\right)

Aditya Kumar - 3 years, 6 months ago

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Solution to Problem 15:

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

Integrating πcotπz(z2+1)2 \frac{\pi \cot \pi z}{(z^2+1)^2} around the positively-oriented square contour CNC_N with corners at ±(N+12)±i(N+12)\pm(N+\tfrac12) \pm i(N+\tfrac12), we deduce that 12πiCNπcotπz(z2+1)2dz=(n=NNResz=n+Resz=i+Resz=i)πcotπz(z2+1)2dz=1+2n=1N1(n2+1)2+(Resz=i+Resz=i)πcotπz(z2+1)2\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 Resz=±iπcotπz(z2+1)2  =  ddzπcotπz(z±i)2z=±i  =  14π2cosech2π14πcothπ \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 12πiCNπcotπz(z2+1)2dz  =  1+2n=1N1(n2+1)212π2cosech2π12πcothπ \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 CNC_N tends to 00 as NN \to \infty, the result follows.

As before, I pass on setting the next problem.

Mark Hennings - 3 years, 6 months ago

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Sir I can't think of a good problem and Ishan is down with fever. Can you post the next problem here?

Aditya Kumar - 3 years, 6 months ago

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Alternate Method:

n=11(n2+1)2=14n=1{1(1+in)211+in+1(1+in)2+11+in}\displaystyle \sum_{n=1}^\infty \frac{1}{(n^2+1)^2}=\frac{1}{4}\sum_{n=1}^\infty \left\{ \frac{1}{(-1+in)^2}-\frac{1}{-1+in}+\frac{1}{(1+in)^2}+\frac{1}{1+in}\right\}

Here, we can use polygamma function and use their reflection formulas to get to the final result.

Aditya Kumar - 3 years, 6 months ago

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Problem 14:

Prove that:

n=1ψ1(n)2nn2=19π41440π224log2(2)log4(2)24ζ(3)log(2)4\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2}=\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{24}-\frac{\zeta(3) \log(2)}{4}

This problem has been solved by Mark Hennings.

Aditya Kumar - 3 years, 6 months ago

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Solution to Problem 14:

We note that n=1ψ(1)(n)2nn2=12ζ(2)n=112n+1(n+1)2[Hn(2)ζ(2)]=ζ(2)n=112nn2n=1Hn(2)2n+1(n+1)2=ζ(2)[112π212ln22]n=1Hn(2)(n+1)22n+1 \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 A(x)=n=1(Hn(2))2xn  =  n=1(Hn1(2)+1n2)2xn=n=1(Hn(2))2xn+1+2n=1Hn(2)(n+1)2xn+1+Li4(x)=xA(x)+2n=1Hn(2)(n+1)2xn+1+Li4(x)(1x)A(x)=2n=1Hn(2)(n+1)2xn+1+Li4(x) \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=12x=\tfrac12, 12n=1(Hn(2))22n  =  2n=1Hn(2)(n+1)22n+1+Li4(12) \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 n=1Hn(2)(n+1)22n+1  =  11440π4124π2ln22+124ln42+14ζ(3)ln2 \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 n=1ψ(1)(n)2nn2  =  191440π4124π2ln22124ln4214ζ(3)ln2 \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.

Mark Hennings - 3 years, 6 months ago

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This question was essentially the same as the previous question. @Aditya Kumar If you set the next question, please post a new one.

Ishan Singh - 3 years, 6 months ago

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@Ishan Singh Can you post the problem? Post it in this note as it will be the 15th problem (don't count spl prob).

Aditya Kumar - 3 years, 6 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, 6 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} Now (differentiate back to check): 0xLi2(v)1vdv=13π2ln(1x)+ln2(1x)lnx+ln(1x)Li2(x)+2Li3(1x)2ζ(3)0xLi2(v)1vlnvdv=145π416π2ln2(1x)+14ln4(1x)+12ln2(1x)ln2x+(ln2(1x)ln(1x)lnx+ln2x)Li2(x)12Li2(x)2+ln2(x1x)Li2(x1x)+2ln(1x)Li3(1x)2lnxLi3(x)+2ln(1x)Li3(x1x)2lnxLi3(x1x)2Li4(1x)+2Li4(x)+2Li4(x1x) \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(12)  =  11440π4124π2ln22+124ln42+14ζ(3)ln2 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 A(x)=n=1(Hn(2))2xn  =  n=1(Hn1(2)+1n2)2xn=n=1n=1(Hn(2))2xn+1+2n=1Hn(2)(n+1)2xn+1+Li4(x)=xA(x)+2B(x)+Li4(x) \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 n=1(Hn(2))22n=A(12)  =  4B(12)+2Li4(12)=1360π416π2ln22+16ln42+2Li4(12)+ζ(3)ln2 \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.

Mark Hennings - 3 years, 6 months ago

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Can you please help me with this query which is related to the above integrals?

Ishan Singh - 3 years, 6 months ago

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The main integral to evaluate here is ln2(1x)lnxxdx \displaystyle \int \dfrac{\ln^2(1-x) \ln x}{x} \mathrm{d}x. Rest of the integrals follow from this using Euler's Reflection Formula and/or IBP. The integral Li2(x)1xdx \displaystyle \int \dfrac{\operatorname{Li}_{2}(x)}{1-x} \mathrm{d}x is evaluated in my solution of Problem 2.

Ishan Singh - 3 years, 6 months ago

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I was wondering whether we can extend the result to other weights too?

Ishan Singh - 3 years, 6 months ago

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I have solved the indefinite integrals by using Landen's Identity.

Ishan Singh - 3 years, 6 months ago

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Solution to Problem 13:

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

f(x)=n=1(Hn1(2)+1n2)2xnf\left( x \right) =\sum _{ n=1 }^{ \infty }{ { \left( { H }_{ n-1 }^{ (2) }+\frac { 1 }{ { n }^{ 2 } } \right) }^{ 2 }{ x }^{ n } }

f(x)=n=1(Hn1(2))2xn+Li4(x)+2n=1Hn1(2)xnn2f\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 nn to n+1n+1, we get:

f(x)=xf(x)+Li4(x)+2n=1Hn(2)xn+1(n+1)2(...1)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 n=1Hn(2)xn+1(n+1)2\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) }{ x }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } }

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

I'll use the identity: n=1Hn(2)xnn=0xLi2(t)t(1t)dt\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:

n=1Hn(2)xn1=Li2(x)x(1x)\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:

n=1Hn(2)xn=Li2(x)(1x)\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. xx from 0 to x:

n=1Hn(2)xn+1n+1=2Li3(1x)2Li2(1x)ln(1x)Li2(x)ln(1x)ln(x)(ln(1x))22ζ(3)\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 xx:

n=1Hn(2)xnn+1=2Li3(1x)x2Li2(1x)ln(1x)xLi2(x)ln(1x)xln(x)(ln(1x))2x2ζ(3)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 22, we get:

n=1Hn(2)xn+1(n+1)2=0x[2Li3(1x)x2Li2(1x)ln(1x)xLi2(x)ln(1x)xln(x)(ln(1x))2x2ζ(3)x]dx\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 11, we get:

f(x)=Li4(x)1x+21x0x[2Li3(1x)x2Li2(1x)ln(1x)xLi2(x)ln(1x)xln(x)(ln(1x))2x2ζ(3)x]dxf\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=12x=\frac { 1 }{ 2 } and compute the integral. Then we finally get:

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

Aditya Kumar - 3 years, 6 months ago

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I'm impressed. The last integral is a brute.

Mark Hennings - 3 years, 6 months ago

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@Mark Hennings In fact we can even generalize the result by finding the indefinite integral of the last one and in turn the generating function n=1(Hn(2))2xn \displaystyle \sum_{n=1}^{\infty} (H_{n} ^{(2)})^2 x^n . The solution gets too long though.

Ishan Singh - 3 years, 6 months ago

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@Ishan Singh very beautiful. we need to be very very cautious.

Srikanth Tupurani - 10 months, 3 weeks 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, 6 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, 6 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, 6 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, 6 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, 6 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, 6 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, 6 months ago

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Problem 10:

Evaluate: k=1ζ(2k+1)12k+3\displaystyle \sum_{k=1}^\infty\frac{\zeta(2k+1)-1}{2k+3}

This problem has been solved by Mark Hennings.

Aditya Kumar - 3 years, 6 months ago

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Solution To Problem 10 :

From the generating function of Riemann Zeta, we have,

n=1ζ(n+1) xn=γψ(1x) \sum_{n=1}^{\infty} \zeta(n+1) \ x^{n} = - \gamma - \psi(1-x)