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Timeline for Has anyone seen this series?

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Mar 23, 2014 at 19:04 comment added Lucian Glad to hear that, @Anand! :-) Though now I realize that the same approximation could also have been derived from Stirling's formula.
Mar 23, 2014 at 15:20 vote accept Anand
Mar 22, 2014 at 9:44 comment added მამუკა ჯიბლაძე Well you can also substitute $t=e^x$ and then integrate 1/4 times wrt $x$...
Mar 22, 2014 at 5:50 comment added Lucian Maximum relative error of about $10-15$% at $t=0$, decreasing exponentially towards $0$ as t increases.
Mar 22, 2014 at 5:02 comment added Anand Thanks Lucian! Do you know how good is this approximation? Good upper and lower bounds will also be great! I will do some calculations tomorrow. Thanks a lot!
Mar 22, 2014 at 4:42 comment added Lucian $\displaystyle\sum_1^\infty\frac{t^n}{n!~n^a}{\large\approx} \sum_1^\infty\frac{t^n}{(n+a)!}$ for $a=\dfrac14$
Mar 22, 2014 at 3:37 comment added Anand If you interpret your $(\int)^{1/4}$ by the fractional integral $J^{1/4}$ as in en.wikipedia.org/wiki/Fractional_calculus, what your get for $t^n$ is $J^{1/4} t^n = \frac{t^{n+1/4}n!}{\Gamma(n+1/4)}$. We cannot get $n^{1/4}$ in the denominator.
Mar 22, 2014 at 3:22 comment added Lucian 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.
Mar 22, 2014 at 2:52 comment added Anand 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!
Mar 22, 2014 at 1:19 history answered Lucian CC BY-SA 3.0