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### 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? $$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$ ($H_n=\sum_{k=1}^n\frac{1}{k}$).
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### 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 $S=\lim\limits_{k\to\infty}s_k$ ...
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### 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 ...
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### 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 ...
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 $$\frac{(H_n)^2}{n}\... 3answers 732 views ### How to compute \sum_{n=1}^\infty\frac{H_n^2}{n^32^n}? Can we evaluate \displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n} ? where H_n=\sum_{k=1}^n\frac1n is the harmonic number. A related integral is \displaystyle\int_0^1\frac{\ln^2(1-x)\... 2answers 632 views ### Two powerful alternating sums \sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2} and \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2} where H_n is the harmonic number and can be defined as: H_n=1+\frac12+\frac13+...+\frac1n H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2} these two sums are already solved by Cornel using ... 3answers 153 views ### Proving \sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a) Mathematica gives$$\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$$All I did is reindexing then using the series property \sum_{n=... 1answer 253 views ### How can I evaluate \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx I am trying to evaluate \displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx I first tried using the series expansion for the dilogarithm like this$$\sum _{n=1}^{\...
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)^...