I come across the following infinite series.

$$ \sum_{n=1}^{\infty} \frac{t^n}{n!\: n^{a}}, \quad\text{for $t>0$ and $a>0$}. $$

In particular, I am interested in the case where $a=1/4$.

Thanks for any hints and references!


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    $\begingroup$ I somehow suspected a "Wanted dead or alive" like poster when I first read the title... $\endgroup$ – Johannes Hahn Mar 22 '14 at 18:40

For natural values of a, take $\displaystyle\frac{e^t-1}t=\sum_1^\infty\dfrac{t^{n-1}}{n!}$, then apply the operator $\bigg(\displaystyle\frac1t\cdot\int\bigg)$ a times to it. For $a\not\in\mathbb N$, such as $a=\dfrac14$, welcome to the “marvelous” world of fractional calculus and Riemann-Liouville integrals.

For natural values of a, the series can also be expressed in terms of the (generalized) hypergeometric function $_{a+1}{\large F}_{a+1}\bigg(\underbrace{1,1,\ldots,1}_{a+1}~;~\underbrace{2,2,\ldots,2}_{a+1}~;~t\bigg)$.

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  • $\begingroup$ Thanks Lucian for your answer. I am interested in the case where a is between 0 and 1. I don't see how the fractional calculus can bring $n^a$ to the denominator, rather some Gamma functions. Could you please give more hints on this case? Thanks a lot! $\endgroup$ – Anand Mar 22 '14 at 2:52
  • $\begingroup$ For $a\in(0,1)$, squeeze it in between $f_0(t)=e^t-1$ and $f_1(t)=Ei(t)-\ln t-\gamma$. See exponential integral. Also, if $f_a(t)=t\cdot\bigg(\dfrac1t\displaystyle\int\bigg)^a\circ \bigg(\dfrac{e^t-1}t\bigg)$ for natural a, then I'm assuming the same should also hold true for $a\not\in\mathbb N$. I'd be surprised if it wouldn't. $\endgroup$ – Lucian Mar 22 '14 at 3:22
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    $\begingroup$ $\displaystyle\sum_1^\infty\frac{t^n}{n!~n^a}{\large\approx} \sum_1^\infty\frac{t^n}{(n+a)!}$ for $a=\dfrac14$ $\endgroup$ – Lucian Mar 22 '14 at 4:42
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    $\begingroup$ Maximum relative error of about $10-15$% at $t=0$, decreasing exponentially towards $0$ as t increases. $\endgroup$ – Lucian Mar 22 '14 at 5:50
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    $\begingroup$ Glad to hear that, @Anand! :-) Though now I realize that the same approximation could also have been derived from Stirling's formula. $\endgroup$ – Lucian Mar 23 '14 at 19:04

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