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!
Anand
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9$\begingroup$ I somehow suspected a "Wanted dead or alive" like poster when I first read the title... $\endgroup$– Johannes HahnCommented Mar 22, 2014 at 18:40
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1 Answer
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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! $\endgroup$– AnandCommented Mar 22, 2014 at 2:52
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$\begingroup$ 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. $\endgroup$– LucianCommented Mar 22, 2014 at 3:22
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1$\begingroup$ $\displaystyle\sum_1^\infty\frac{t^n}{n!~n^a}{\large\approx} \sum_1^\infty\frac{t^n}{(n+a)!}$ for $a=\dfrac14$ $\endgroup$– LucianCommented Mar 22, 2014 at 4:42
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1$\begingroup$ Maximum relative error of about $10-15$% at $t=0$, decreasing exponentially towards $0$ as t increases. $\endgroup$– LucianCommented Mar 22, 2014 at 5:50
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1$\begingroup$ Glad to hear that, @Anand! :-) Though now I realize that the same approximation could also have been derived from Stirling's formula. $\endgroup$– LucianCommented Mar 23, 2014 at 19:04