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I´ve got the following infinite sum which - determined by the ratio test - is convergent for $\{p\in\mathbb{R}\ |\ p>1\}$

$$\left(\frac{1}{2\sqrt{\pi}}\right)\cdot\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p^{2n-1}}\cdot\frac{1}{\sqrt{n-1}}.$$

This series is a result of applying Stirling's approximation to the sum of an infinite series given by (that's why the whole sum is multiplied by a constant):

$$\frac{1}{2}\cdot\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p}\cdot\frac{1}{p^{2n-2}}\cdot\frac{(2n-2)!}{(n-1)!(n-1)!}$$

Which can be expanded to:

$$2\sum\limits_{n=1}^\infty \frac{(2p-1)^{n-1}}{p}\cdot\frac{1}{p^{2n-2}}\cdot\left(\frac{(2n-3)}{(2n-2)}\cdot\frac{(2n-5)}{(2n-4)}\cdot\frac{(2n-7)}{(2n-6)}\cdot\frac{(2n-9)}{(2n-8)}\ \cdots\ \frac{3}{2}\right)$$

My question is the following: how would I go about approximating - if it is even possible to find the exact solution - either this series or its Stirling's approximate? And assuming it's impossible: if I was to compute the result, how would I show that the result is correct to a given number of decimal places?

Clayton
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Bruno KM
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  • The term for $n=1$ is $\frac1p\frac1{\sqrt0}$. Did you perhaps mean something else? – robjohn Sep 18 '15 at 22:13
  • Oh, I did mean what I wrote, but I didn't see this issue before. But you could just start the sum at $n=2$ and add the first term separately, before using Stirling's approximation for factorials. – Bruno KM Sep 19 '15 at 10:23

1 Answers1

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Stirling's approximation only applies for large $n$: it certainly doesn't apply for $n-1=0$!

Actually, as written, I don't think you need to take Stirling's approximation for your sum: the coefficients are the central binomial coefficients, $\binom{2n}{n}$; these have generating function $$ \sum_{n=0}^{\infty} \binom{2n}{n} z^n = \frac{1}{\sqrt{1-4z}}, $$ and rewriting your sum, I think it becomes $$ \frac{1}{2p} \sum_{n=0}^{\infty} \left( \frac{2p-1}{p^2} \right)^n \binom{2n}{n} = \frac{1}{2p} \left( 1-4\frac{2p-1}{p^2} \right)^{-1/2} = \frac{1}{2\sqrt{p^2-8p+4}} $$ when $p>0$. However, we also need $z=\frac{2p-1}{p^2}<\frac{1}{4}$, which reduces to $p>2(2+\sqrt{3})$ or $p<2(2-\sqrt{3})$.

Community
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Chappers
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  • (+1) of course the original series has a problem for $n=1$ (division by zero). – robjohn Sep 18 '15 at 22:27
  • The Stirling's Approximation does have the problem of division by zero for $n=1$ , not the original series. It is true, however, that I should remove at least the first term from the sum and add it separately. – Bruno KM Sep 19 '15 at 10:26
  • I want to sincerely thank you for the answer Chappers, it helps a lot. Nevertheless, could you also elaborate on how you got these results, or link sources relevant to proving your suggestion? – Bruno KM Sep 19 '15 at 10:34
  • @Kasper See this question, for example. I'll link it in the answer, too. – Chappers Sep 19 '15 at 12:26