Linked Questions

44
votes
9answers
4k views

Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$

How can I find a closed form for the following sum? $$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$ ($H_n=\sum_{k=1}^n\frac{1}{k}$).
30
votes
5answers
942 views

How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?

I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$, where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$. Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ ...
30
votes
5answers
964 views

Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$?

I am seeking a closed-form solution for this double sum: \begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?. \end{eqnarray*} I will turn it into $3$ tough ...
16
votes
3answers
738 views

How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?

Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ? where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number. A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\...
15
votes
5answers
888 views

Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $

I proved the following result $$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$ After ...
12
votes
5answers
591 views

Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$

Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$? Quick Results This series converges because $$\frac{(H_n)^2}{n}\...
9
votes
2answers
638 views

Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$ these two sums are already solved by Cornel using ...
4
votes
3answers
159 views

Proving $\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a)$

Mathematica gives $$\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a),\quad 0<a<1$$ All I did is reindexing then using the series property $\sum_{n=...
3
votes
1answer
262 views

How can I evaluate $\int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$

I am trying to evaluate $\displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$ I first tried using the series expansion for the dilogarithm like this $$\sum _{n=1}^{\...
1
vote
1answer
129 views

An equality about infinite sums involving squared harmonic numbers [duplicate]

I found the following equality. But I forgot the source as well as the solution. If you give the proof, I appreciate it. $$\sum_{k=1}^\infty \frac{1}{k^2}\left( 1 + \frac12 + \dots + \frac1k \right)^...