All Questions
Tagged with polylogarithm harmonic-numbers
99
questions
3
votes
1
answer
308
views
Find the series expansion of $\frac{\ln^4(1-x)}{1-x}$
How to prove that
$$\frac{\ln^4(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}\right)x^n=S_n$$
where $H_n^{(a)}=\sum_{k=1}^n\frac1{k^a}$...
- 25.8k
3
votes
2
answers
282
views
Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$
How to evaluate the following challenging sum:
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+...
- 25.8k
4
votes
1
answer
439
views
Advanced Sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}$
How to prove
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}=
\\ \small{\frac43\ln^32\zeta(2)-\frac72\ln^22\zeta(3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64}\zeta(5)-\frac4{15}\ln^52-8\ln2\...
- 25.8k
4
votes
1
answer
278
views
Is there a closed form for the integral $ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $
Is there a closed form for the integral
$$ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $$
where $ \displaystyle \mathrm{Li}_2(x) \;\;=\;\; \sum_{n=1}^\infty \frac{x^n}{n^2} $
a ...
- 496
9
votes
5
answers
2k
views
A group of important generating functions involving harmonic number.
How to prove the following identities:
$$\small{\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)...
- 25.8k
7
votes
1
answer
457
views
Compute $\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}$
How to prove that
$$\small{\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}=\frac1{12}\ln^52+\frac{13}{128}\zeta(5)-\frac12\ln^32\zeta(2)+\frac74\ln^22\zeta(3)-\frac{17}{8}\ln2\zeta(4)+2\ln2\...
- 25.8k
8
votes
1
answer
464
views
Very advanced sums: Compute $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2}$ and $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2}$
How to prove:
$$S_1=\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2}
=\frac{23 }{32}\zeta (2) \zeta (3)-\frac{581}{128} \zeta (5)-\frac{2}{3}\ln ^32 \zeta (2)+\frac{7}{4} \ln^22\zeta (3)\\ +\...
- 25.8k
9
votes
3
answers
776
views
Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$
Prove that
$$S=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$
where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the n$th$ generalized harmonic ...
- 25.8k
4
votes
1
answer
173
views
Prove $\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)$
How to prove, in a simple way that
$$I=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)\ ?$$
Where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the ...
- 25.8k
16
votes
3
answers
918
views
How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?
Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ?
where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number.
A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\...
- 25.8k
10
votes
1
answer
521
views
Mind-blowing Sums: Compute $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}$ and $\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}$
How to prove the following two sums
\begin{align}
\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}&=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}...
- 25.8k
4
votes
1
answer
282
views
Compute $2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}$
How to prove that
$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}=7\zeta(7)+\frac{7}{4}\zeta(3)\zeta(4)-\frac32\zeta(2)\zeta(5)\tag{1}$$
where $H_n^{(p)}=1+...
- 25.8k
10
votes
1
answer
385
views
Challenging sum: Calculate $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^52^k{2k \choose k}}$
We proved in our previous solutions here and here the following two sums:
$$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=\frac1{4}\zeta(3)-\frac1{6}\ln^32$$
$$\sum_{k=1}^\infty\frac{(-1)^...
- 25.8k
14
votes
4
answers
2k
views
Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$
Prove that
I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean:
The ...
- 25.8k
10
votes
1
answer
991
views
Computing $\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$ or $\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$
Challenging Integral:
\begin{align}
I=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx&=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{...
- 25.8k