n1=1n2=1nk=1(1)n1+n2++nkHn1+n2++nkn1n2n3nk\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}

n1=1n2=1nk=1(1)n1+n2++nkHn1+n2++nkn1n2n3nk=(1)kn=1k(kn)(log2)kn[(1)nn!ζ(n+1)+2(1)n1r=0n(nr)r!(log2)nrLin+1(12)]\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,

Hn=011xn1x  dx\displaystyle H_n=\int_0^1 \dfrac{1-x^n}{1-x}\; dx

Substituting this into our sum SS we have,

S=01n1,n2,,nk=1(1)n1+n2++nkn1n2nkn1,n2,,nk=1(1)n1+n2++nkxn1+n2++nkn1n2nk1x  dx=(1)k01logk2logk(1+x)1x  dx=(1)k1/21logk2logk(2x)1x  dx=(1)kn=1k(kn)(log2)kn1/21lognx1x  dx\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 ,

1/21lognx1x  dx\displaystyle \int_{1/2}^1 \frac{\log^n x}{1-x}\; dx

To find this we evaluate a preliminary integral,

1/21xm  dx=1m+12m1+m1/21xmlognx  dx=dndmn[1m+12m1+m]1/21xmlognx  dx=(1)nn!(m+1)n+1+2m(1)n1r=0n(nr)(log2)nrr!(m+1)r+1\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,

1/21lognx1x  dx=m=01/21xmlognx  dx=m=0[(1)nn!(m+1)n+1+2m(1)n1r=0n(nr)(log2)nrr!(m+1)r+1]=(1)nn!ζ(n+1)+2(1)n1r=0n(nr)r!(log2)nrLin+1(12)\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,

S=(1)kn=1k(kn)(log2)kn[(1)nn!ζ(n+1)+2(1)n1r=0n(nr)r!(log2)nrLin+1(12)]\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]

Note by Aditya Narayan Sharma
2 years, 4 months ago

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What? :/

Steven Jim - 2 years, 4 months ago

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

Sathvik Acharya - 2 years, 4 months ago

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

Steven Jim - 2 years, 4 months ago

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

Aditya Narayan Sharma - 2 years, 2 months ago

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

Steven Jim - 2 years, 2 months ago

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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

Md Zuhair - 2 years, 1 month ago

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

Sathvik Acharya - 2 years, 1 month ago

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

Aaghaz Mahajan - 7 months, 3 weeks ago

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

Md Zuhair - 7 months, 3 weeks ago

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

Aaghaz Mahajan - 7 months, 3 weeks ago

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@Aaghaz Mahajan lets see.. thanks

Md Zuhair - 7 months, 3 weeks ago

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@Aaghaz Mahajan bhai mera 2 marks se KVPY choot gaya

Md Zuhair - 7 months ago

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@Md Zuhair 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)

Aaghaz Mahajan - 6 months, 4 weeks ago

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

Md Zuhair - 6 months, 4 weeks ago

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Very nice!

Hunter Edwards - 2 years ago

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