Skip to main content

Questions tagged [euler-sums]

For questions about and related to so-called Euler Sums, that are sums having [tag:harmonic-numbers] and negative integer powers of the index as coefficients. DO NOT USE THIS TAG for Euler summation.

7 votes
2 answers
195 views

how to find the following series: $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j ...
Wessel's user avatar
  • 73
8 votes
1 answer
487 views

I've been interested in the Basel problem and its famous solution $$ \sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}. $$ Recently I saw this video along with a comment (highlighted from the link) ...
FishDrowned's user avatar
  • 1,051
9 votes
2 answers
755 views

Question How to evaluate $$ \int_0^1 \frac{\ln(1-x) \ln^2 x \ln^2(1+x)}{1-x} \, dx $$ My attempt To evaluate the integral $$ I = \int_0^1 \frac{\ln(1-x) \ln^2 x \ln^2(1+x)}{1-x} \, dx $$ We can ...
Martin.s's user avatar
  • 8,206
9 votes
1 answer
512 views

The following problem $$\sum_{n=1}^\infty\frac{2^{2n+1}H_n}{n(2n+1)^2\binom{2n}{n}}+\sum_{n=1}^\infty\frac{(H_{n-1})^2}{(2n-1)^2 2^{2n}}\binom{2n}{n}+\sum_{n=1}^\infty\frac{H^{(2)}_{n-1}}{(2n-1)^2 2^{...
Ali Olaikhan's user avatar
  • 28.2k
3 votes
2 answers
166 views

Prove $ S_{p,q} = \zeta(p+q) - \frac{(-1)^q}{\Gamma(q)} \displaystyle\int_{0}^1(\log{t})^{q-1}\operatorname{Li}_p(t)\frac{1}{1-t}\mathrm{d}t $, where $S_{p,q}$ denotes the linear Euler sum $\...
tys's user avatar
  • 512
3 votes
2 answers
133 views

Show that $$\int^1_0 x^{m-1}\ln^2(1-x)dx=\frac{2}{m}\sum_{k=1}^m\frac{H_{k}}{k}$$ We know: $$\ln(1-x)=-\sum^{\infty}_{n=1}\frac{x^n}{n}$$ and using cauchy product Therfore: $$\begin{align*} \ln^2(1-x)&...
epsilon's user avatar
  • 3,156
3 votes
1 answer
209 views

For the past couple of days I have been looking at Euler Sums, and I happened upon this particular one: $$ \sum_{n=1}^{\infty}\left(-1\right)^{n}\, \frac{H_{n}}{n^{5}} $$ I think most people realize ...
Jessie Christian's user avatar
9 votes
1 answer
427 views

I was trying to evaluate this famous integral $$\int_{0}^{1} \frac{\ln (x) \ln^{2}(1+x) \ln(1-x)}{x} \ dx $$ Here is my attempt so solve the integral \begin{align} &\int_{0}^{1} \frac{\ln (x) \ln^{...
user avatar
5 votes
1 answer
337 views

Question: closed form of $$\int_{[0 ; 1]} \frac{\operatorname{Li}_3\left(-x^2\right)}{1+x} d x$$ My try to solve the integral $$ \begin{aligned} & I=\int_{[0 ; 1]} \frac{\operatorname{Li}_3\left(-...
user avatar
7 votes
0 answers
282 views

How to integrate $$\displaystyle{\Large\displaystyle \int_0^1\frac{\ln^3(1+x)\,\ln^3x}x\mathrm{d}x}$$ My idea \begin{align*} \int_{0}^{1} \frac{\ln^3 (1+x) \ln^3 x}{x} \, \mathrm{d}x &= -18 \sum_{...
Martin.s's user avatar
  • 8,206
3 votes
2 answers
332 views

How to calculate this sum $$\sum_{n=1}^{\infty} \frac{H_n \cdot H_{n+1}}{(n+1)(n+2)}$$ Attempt The series telescopes. We have $$=\frac{H_n \cdot H_{n+1}}{(n+1)(n+2)} = \frac{H_n \cdot H_{n+1}}{n+1} - ...
Martin.s's user avatar
  • 8,206
1 vote
2 answers
126 views

$$\sum_{n=1}^{\infty} (-1)^{n+1} H_n \left( \frac{1}{n+1} - \frac{1}{n+3} + \frac{1}{n+5} - \ldots \right) = \frac{\pi}{16} \cdot \log(2) + \frac{3}{16} \cdot \log(2) - \frac{\pi^2}{192}$$ $$\sum_{k=...
Martin.s's user avatar
  • 8,206
0 votes
0 answers
95 views

Recently I tried to estimate this sum but without much success, would you have any methods to suggest? $S=\sum_{n=1}^{+\infty} (-1)^n \frac{H_{2n}}{n^2}$
azur's user avatar
  • 167
2 votes
0 answers
102 views

$$\sum_{n=1}^{\infty}\frac{\left(16(-1)^n+5\right)\left(\phi H_n+\frac{1}{n^3}\right)+\frac{11}{n^3}}{n^2}=\zeta(5)\tag1$$ $\phi=\frac{1+\sqrt{5}}{2}$ $H_n=\sum_{k=1}^{n}\frac{1}{k}$ we expanded $(1)$ ...
Sibawayh's user avatar
  • 1,493
3 votes
0 answers
152 views

I need a closed form of the sum $$\sum_{n=1}^{\infty}\frac{((n-1)!)^2(H_{2n}-H_{n-1})}{(2n)!} $$ where $H_n$ denotes the Harmonic number and "!" is the factorial notation. I tried: Harmonic ...
Max's user avatar
  • 1,059

15 30 50 per page