$\displaystyle \begin{aligned} &\sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty (-1)^{n_1+n_2+\cdots+n_k}\dfrac{H_{n_1+n_2+\cdots+n_k}}{n_1 n_2 n_3 \cdots n_k} \\ &= (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \left[(-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \right]\end{aligned}$

The above is the claim which is proved here after in this note.

**Proof:**

First note that,

$\displaystyle H_n=\int_0^1 \dfrac{1-x^n}{1-x}\; dx$

Substituting this into our sum $S$ we have,

$\displaystyle \begin{aligned} S&= \int_0^1 \dfrac{\sum_{n_1,n_2,\cdots,n_k=1}^\infty \dfrac{(-1)^{n_1+n_2+\cdots+n_k}}{n_1 n_2\cdots n_k}-\sum_{n_1,n_2,\cdots,n_k=1}^\infty \dfrac{(-1)^{n_1+n_2+\cdots+n_k}x^{n_1+n_2+\cdots+n_k}}{n_1 n_2\cdots n_k}}{1-x}\; dx \\ &= (-1)^k \int_0^1 \dfrac{\log^k 2-\log^k (1+x)}{1-x}\; dx \\ &= (-1)^k \int_{1/2}^1 \dfrac{\log^k 2-\log^k (2x)}{1-x}\; dx \\ &= (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \int_{1/2}^1 \dfrac{\log^n x}{1-x}\; dx \end{aligned}$

Now it remains to calculate ,

$\displaystyle \int_{1/2}^1 \frac{\log^n x}{1-x}\; dx$

To find this we evaluate a preliminary integral,

$\displaystyle \begin{aligned} & \int_{1/2}^1 x^m \; dx = \dfrac{1}{m+1}-\dfrac{2^{-m}}{1+m} \\ & \int_{1/2}^1 x^m \log^n x \; dx = \dfrac{d^n}{dm^n}\left[\dfrac{1}{m+1}-\dfrac{2^{-m}}{1+m}\right] \\ & \int_{1/2}^1 x^m \log^n x \; dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}+2^{-m}(-1)^{n-1} \sum_{r=0}^n \binom{n}{r}\dfrac{(\log 2)^{n-r}r!}{(m+1)^{r+1}}\end{aligned}$

Now coming back to the main integral,

$\displaystyle \begin{aligned} \int_{1/2}^1 \frac{\log^n x}{1-x}\; dx &= \sum_{m=0}^\infty \int_{1/2}^1 x^m \log^n x \; dx \\ &= \sum_{m=0}^\infty \left[ \dfrac{(-1)^n n!}{(m+1)^{n+1}}+2^{-m}(-1)^{n-1} \sum_{r=0}^n \binom{n}{r}\dfrac{(\log 2)^{n-r}r!}{(m+1)^{r+1}}\right] \\ &= (-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \end{aligned}$

Putting this back,

$\displaystyle S = (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \left[(-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \right]$

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## Comments

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TopNewestWhat? :/

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Same reaction here buddy:) I need to improve my calculus skills

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So do I :) She made me look like a noob :/

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He is so good at calculus that once his name was I guess published at the american mathematics daily becoz he sent a solution to a research level problm

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Great! I guess this something that he found interesting and wanted to share (Though i understand nothing of it)

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WOAH!!! Seriously???! This is soo great!

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P.S. Main to iss saal ke liye prepare kar rhaa hun.....But mera ek dost hai, uska 11th waala clear hogaya and he got AIR 72....!!! Uska to pakka ho jaayega IISC mein......But problem is..........Usse research mein BILKUL interest nhi hai!!!! XD!!!! Pata nhi kyon hote hain aise log.......!!! XD!! (No offense.....:P)

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Very nice!

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