# $\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty (-1)^{n_1+n_2+\cdots+n_k}\dfrac{H_{n_1+n_2+\cdots+n_k}}{n_1 n_2 n_3 \cdots n_k}$

\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,

$\displaystyle S = (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \left[(-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \right]$ 4 years, 2 months ago

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Sort by:

Very nice!

- 3 years, 10 months ago

What? :/

- 4 years, 2 months ago

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

- 3 years, 11 months ago

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

- 2 years, 5 months ago

yep i know.. he is a genius.. he is running a Youtube Channel too at present, Name is ANS Academy

- 2 years, 5 months ago

DAMN!!! Mujhe to uss channel ke baare mein aaj pata chala!!! Btw, if you want some interesting youtube channels, try "Faculty of Khan"....!!

- 2 years, 5 months ago

bhai mera 2 marks se KVPY choot gaya

- 2 years, 4 months ago

Wait shitt!!!!! Abbe yaar......!!!! 2 MARKS??!!!!!! Chal still don't lose hope......tu JEE mein bhi agar top 250 mein rank le aaye to IISC ho jaata hai...!! So, best of luck for it!!!!!

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)

- 2 years, 4 months ago

True.... Mera ek dost hai.. uska bhi same situation. AIR 55 aya .... SX mei .. aur phir IIT mein CSE lega bolta hai

- 2 years, 4 months ago

lets see.. thanks

- 2 years, 5 months ago

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

- 3 years, 11 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago

I hope the 'she' doesn't refer to me , I'm a he :p

Um... I really should learn how to identify he/she, for real XD

- 4 years ago