All Questions
23
questions
1
vote
1
answer
112
views
asymptotic behaviour of polylogarithmic function
I would like to understand the asymptotic behaviour as $a \rightarrow 0$ of the function
$$
f(a) := \sum\limits_{k=2}^{\infty} e^{ - a^2 k}{k^{-3/2}}
$$
More precisely, I would like to obtain an ...
- 200
0
votes
1
answer
93
views
$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$ as polylogarithm
$$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$$
It is very clear for me that it has to be polylogarithm function but as it is partial sum I tried to split the series as
$$\sum_{i=1}^{\infty} \frac {x^{...
- 3
3
votes
0
answers
316
views
Two tough integrals with logarithms and polylogarithms
The following two integrals are given in (Almost) Impossible Integrals, Sums, and Series (see Sect. $\textbf{1.55}$, page $35$),
$$i) \int_0^{\pi/2} \cot (x) \log (\cos (x)) \log ^2(\sin (x)) \...
- 5,495
2
votes
4
answers
157
views
Errors are decreasing in series $\sum_{n=1}^\infty(-1)^n/n^4$?
Let $v=\sum_{n=1}^\infty(-1)^n/n^4$ ($v$ for "value"), let $S=(\sum_{n=1}^m(-1)^n/n^4)_{m\in\mathbb Z_{\ge1}}$ be the partial sums, and let $e=(|S_n-v|)_{n\in\mathbb Z_{\ge1}}$ be the errors....
- 867
6
votes
1
answer
765
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 25.8k
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 25.8k
1
vote
1
answer
111
views
How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?
where $H_n$ denotes the harmonic number.
I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even
$$-\sum_{n\geq 1}H_{2n}^2\...
- 125
2
votes
1
answer
86
views
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?
Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
- 125
1
vote
2
answers
150
views
How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$
I am having a difficult time evaulating
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$
I have tried the following relation:
$$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$
...
user753116
2
votes
1
answer
291
views
How to compute $\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$
How to compute
$$\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$$
where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$ is the dilogarithm function.
...
- 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
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}=
\\ \small{\frac43\ln^32\zeta(2)-\frac72\ln^22\zeta(3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64}\zeta(5)-\frac4{15}\ln^52-8\ln2\...
- 25.8k
6
votes
2
answers
529
views
$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$
This problem was proposed by Cornel and he showed that
$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+...
- 25.8k
11
votes
2
answers
927
views
Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$
where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$
these two sums are already solved by Cornel using ...
- 25.8k
6
votes
1
answer
181
views
Series power function over exponential function
A typical exercise from calculus is to show that any exponential function eventually grows faster than any power function, i.e.
$$ \lim_{k \to \infty} \frac{k^a}{b^k} = 0 \qquad \text{ for } a,b>1.$...
- 161
3
votes
0
answers
106
views
Proving that swapping the order of this summation is justified
I'm unsure if this has been discovered already, but it's heavily related to my current research, particularly to this question of mine (this conjecture was also originally posted at the beginning of ...
- 856