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5
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Evaluate $\sum\limits_{n=0}^\infty\operatorname W(e^{e^{an}})x^n$ with Lambert W function
$\def\W{\operatorname W} \def\Li{\operatorname{Li}} $
Interested by $\sum_\limits{n=1}^\infty\frac{\W(n^2)}{n^2}$, here is an example where Lagrange reversion applies to a Lambert W sum:
$$\W(x)=\ln(...
- 12.5k
1
vote
1
answer
59
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Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the ...
- 99
0
votes
0
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59
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Stirling number series resummation
\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1}
\left(\frac{x}{x-1}\right)^m}{m!}\end{equation}
Does somebody know the result of this resummation?
Note:
$S_m^{(3)} $ belongs to ...
- 99
0
votes
0
answers
82
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General expression of a triangle sequence
\begin{gather*}
\frac{1}{4} \\
\frac{1}{4} \quad \frac{1}{4} \\
\frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\
\frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\
\frac{...
- 99
3
votes
1
answer
308
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Find the series expansion of $\frac{\ln^4(1-x)}{1-x}$
How to prove that
$$\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 $H_n^{(a)}=\sum_{k=1}^n\frac1{k^a}$...
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