I will be grateful for any ideas to solve the series

$$\sum^\infty_{k=0}\frac{x^k z^k}{k!} \frac{\Gamma(1+a+2k)}{\Gamma(2+k)}{}_2F_1(1,1+a+2k;2+k;z)$$

$a$ is a nonegative integer, $z$ and $x$ are real numbers.


Your expression simplifies to $s(k)=(xz)^k/k!*\sum_{n=0}^\infty z^n*\frac{(2k+a+n)!}{(k+n+1)!}$ where $s(k)$ is the sequence you sum from $k=0$ to infinity. Since $a$ is non-negative, $\frac{(2k+a+n)!}{(k+n+1)!k!}>\binom{2k+a+n}{k}$ for $a>1$ and the limit of $\frac{(2k+a+n)!}{(k+n+1)!k!}$ for large $k$ or $n$ goes to infinity in any event, $(xz)^k$ and $z^n$ would have to converge to zero for large $k$ and $n$, meaning $|xz|<1$. I also believe ${}_2F_1$ would converge for small $z$ because as $n$ and $k$ go to infinity, $\frac{(2k+a+n)!}{(k+n+1)!k!}$ diverges, but at a slower rate. That is, if $f(a,k,n)=\frac{(2k+a+n)!}{(k+n+1)!k!}$ and $n$ is large, then $\lim_{n\rightarrow\infty} \frac{f(a,k,n)}{f(a,k,n-1)}=1/k!$ and if $k$ is large, then $\lim_{k\rightarrow\infty} \frac{f(a,k,n)}{f(a,k-1,n)}=4$, so if $|z|<1/4$ and $|xz|<1$ then the series converges. Otherwise, it will diverge. There is more analysis that can be done on the limiting behavior once $x$, $z$, and $a$ are known.

Since many terms may be required in a partial sum to approximate the limit for large $a$ or $|z|$ near $1/4$ or $|xz|$ near 1, you may want to use a series acceleration method in those cases. I recommend using the Improved-Kummer's acceleration method since it is usually very precise after just one iteration and it only requires a rough knowledge of the limiting behavior of the sequence. It looks too complicated to try to integrate it. The only caveat is that $s(k)$ must be monotone decreasing, so first find the value of $k$ for which $s(k)$ is largest and evaluate the partial sum of $s(k)$ at least up to that value of $k$. You probably will want to evaluate the partial sum past that point. If $s(k)$ is your sequence (above) for fixed values of $x$, $z$, and $a$, then you only need to find a cumulative sum function $G(k)$ such that $\lim_{k\rightarrow\infty}\frac{s(k)}{G(k)-G(k-1)}=c$ where $c$ is a non-zero constant. I suspect that a good first choice for $G(k)$ is a polynomial times the exponential $(4xz^2)^k$. Once you determine a $G(k)$, it is very easy to use the Improved-Kummer's method. I give two examples for how to use it on my site, www.greatestintegerfunctionresearch.org. See section 2.2 in the Full Text.

  • $\begingroup$ Thanks Ken, I appreciate your extensive and informative answer. $\endgroup$
    – Remy
    Apr 15 '12 at 21:55

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