# Looking for the limit of a sum

Looking for a limit, with $1<\alpha\leq 2$, $\sigma>0$:

$$\lim_{p\to \infty } \, \sum _{k=1}^p \left( \frac{1}{2 \pi k!}\left(1+i \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha } \left(1-i \tan \left(\frac{\pi \alpha }{2}\right)\right)^{k/\alpha }+(-1)^k \left(1-i \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha } \left(1+i \tan \left(\frac{\pi \alpha }{2}\right)\right)^{k/\alpha }\right) \left(i^k K^k \sigma ^{1-k} \sec ^2\left(\frac{\pi \alpha }{2}\right)^{-\frac{k}{\alpha }} \Gamma \left(\frac{k+\alpha -1}{\alpha }\right)\right),$$ which converges numerically. It is easy to get acceptable results with special functions for known values of $\alpha$, say with $\alpha=\frac{3}{2}$: $$-\frac{e^{\frac{K^3}{27 \sigma ^3}} \left(\sqrt[3]{2} K^2 \, _0\tilde{F}_1\left(;\frac{4}{3};\frac{K^6}{2916 \sigma ^6}\right)-18 \sigma ^2 \, _0\tilde{F}_1\left(;\frac{2}{3};\frac{K^6}{2916 \sigma ^6}\right)\right)}{9\ 2^{2/3} \sigma }$$

(where $_0\tilde{F}_1$ is the confluent hypergeometric function, $_0F_1(;a;z)=\sum _{k=0}^{\infty } \frac{z^k}{k! (a)_k}$, $(a)_k$ is the Pochhammer function: $(a)_n=a (a+1) \ldots (a+n-1).$ )

There are problems where simplifications of an integral are only known for rational $\alpha$s and, for practical purposes, where $\alpha = \text{Rational}[a, b]$ has $a$ and $b$ being relatively "small" integers. For example, closed form expressions for stable PDFs are available in terms of Meijer G-functions, but only for rational values of $\alpha$. If these summations arise from an application involving stable distributions, then you may be able to get traction by assuming a rational $\alpha$ and applying a simplification downstream in the computation from your result above.
• Thanks a million Bob. This is a discrete summation so I may be able to play with $\alpha=Rational[a,b]$. – Nero Jan 10 '15 at 11:47