MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find a closed form solution for the two sums given by

$$\sum^n_{k=0}\frac{y^k}{k!}(-a+n)_k \left(\frac{2}{z} \right)^k {}_2F_3\left({-\frac{k}{2}, \frac{1 - k}{2}}; {-k, -a+n,1+a-k-n}; z^2\right) K_{-a+n}(z) $$ $$\sum^\infty_{k=0}\frac{y^k}{k!}(-a+n)_k \left(\frac{2}{z} \right)^k {}_2F_3\left({-\frac{k}{2}, \frac{1 - k}{2}}; {-k, -a+n,1+a-k-n}; z^2\right) K_{-a+n}(z) $$

given that $a$ is a nonnegative integer, and $z$ is a nonzero nonegative real number and $K$ is the modified bessel function of the second type.

I found that the value of the hypergeometric series ${}_2F_3\left({-\frac{k}{2}, \frac{1 - k}{2}}; {-k, -a+n,1+a-k-n}; z^2\right)$ is simply a finite sum of the first $k$ terms of ${}_2F_3$. However, If I do that I will a have double sum of the form $\sum^n_{k=0}\sum^k_{q=0}$.

I need to find a closed form for the first finite sum because I have another sum in the form $\sum^\infty_{n=0}$, and therefore, I have to get rid of the first sum from $k$ up to $n$.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.