All Questions
7
questions
3
votes
3
answers
386
views
$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum
Working on the same lines as
This/This and
This
I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$:
$$\operatorname{Li}_{2} \left(\frac{1}{e^{\...
- 862
0
votes
0
answers
96
views
$\operatorname{Li}_{2} \left(\frac12 \right)$ vs $\operatorname{Li}_{2} \left(-\frac12 \right)$ : some long summation expressions
Throughout this post, $\operatorname{Li}_{2}(x)$ refers to Dilogarithm.
While playing with some Fourier Transforms, I came up with the following expressions:
$$2 \operatorname{Li}_{2}\left(\frac12 \...
- 862
0
votes
1
answer
99
views
Converting a 2D lattice sum into a sum over 1D lattice sums in a circle
I'm working on a physics problem. I have a lattice sum, which in 1D is a sum over a linear chain. It reads
$$
f_k = \sum_{n=1}^\infty \frac{2\cos(kn)}{n^3}.
$$
This can be written in terms of ...
- 510
2
votes
0
answers
75
views
The summation $\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi$ [duplicate]
In a recent post
Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$
I evaluated the required ...
- 43.6k
1
vote
1
answer
59
views
Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the ...
- 99
0
votes
0
answers
59
views
Stirling number series resummation
\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1}
\left(\frac{x}{x-1}\right)^m}{m!}\end{equation}
Does somebody know the result of this resummation?
Note:
$S_m^{(3)} $ belongs to ...
- 99
0
votes
0
answers
82
views
General expression of a triangle sequence
\begin{gather*}
\frac{1}{4} \\
\frac{1}{4} \quad \frac{1}{4} \\
\frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\
\frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\
\frac{...
- 99