# Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$

I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$

This is similar to the series $$\frac{1}{4} + \frac{1\cdot 3}{4 \cdot 6} + \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \dotsb,$$which one can show is convergent by Raabe's test, as follows: $$\frac{a_{n+1}}{a_n}= \frac{2n-1}{2n+2}= \frac{2n + 2 - 3}{2n+2}=1 - \frac{3}{2(n+1)}.$$

However, in the series I am looking at, $\frac{a_{n+1}}{a_n}=\frac{(2n-1)^2}{(2n)^2}= 1 - \frac{4n-1}{4n^2}$, so Raabe's test doesn't work (this calculation also happens to show that the ratio and root tests will not work).

Any ideas for this one? It seems the only thing left is comparison...

• Are you sure that the second series converges? How did you apply Raabe's test? Jul 3, 2013 at 23:09
• I'll add that to the question. Jul 3, 2013 at 23:10
• Oh, I thought the terms in the first was square of the second, but it's not.. Jul 3, 2013 at 23:15
• The first sum diverges, it can be done by comparison with $c/n$ for some positive $c$. Jul 3, 2013 at 23:17
• In your serie, shouldn't it be $\frac{1 \cdot 9 \cdot 25 \cdot 49}{4 \cdot 16 \cdot 36 \cdot 64}$ instead of $\frac{1 \cdot 9 \cdot 25 \cdot 36}{4 \cdot 16 \cdot 36 \cdot 64}$ ? Jul 4, 2013 at 9:40

## 4 Answers

$$2\cdot 4\cdots (2n)=2^n n!,\quad 1\cdot 3\cdots (2n-1)=\frac{(2n)!}{2^nn!}$$

By Stirling:

$$\frac{1^2\cdot 3^2\cdots (2n-1)^2}{2^2\cdot 4^2\cdots (2n)^2}=\frac{1}{16^n}\left(\frac{(2n)!}{n!^2}\right)^2\sim \frac{1}{16^n}\left[\frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^{2n}2 \pi n}\right]^2=\frac{1}{\pi n}$$

Hence it diverges.

• I am just learning to use Stirling's formula, and I'm not quite clear how you applied it here. Could you clarify a bit? If we start with $n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}$, how do we obtain the final manipulation that you've done? Thanks Jul 3, 2013 at 23:38
• Use the formula for $(2n)!$ and $(n!)^2$. Jul 3, 2013 at 23:45
• Wow, cool! (extra characters) Jul 4, 2013 at 0:25
• $\frac1\pi\gt\frac{e^{-1/2}}{2}$, but not by much. It is interesting to see how close one can get without resorting to Stirling. :-) of course, $\frac1{\pi n}$ is asymptotic, whereas $\frac{e^{-1/2}}{2n+1}$ is a lower bound.
– robjohn
Jul 4, 2013 at 0:38

There are several methods as commented already, but I will put mine for what is worth.

Here's my answer: Let $$a_n=\frac 1 2 \cdot \frac 3 4 \cdots \frac{2n-1}{2n}$$ Then $$a_n^2=\frac 1 2 \frac 1 2 \frac 3 4 \frac 3 4 \cdots \frac{2n-1}{2n}\frac{2n-1}{2n}$$

$$a_n^2 > \frac 1 2 \frac 1 2 \frac 2 3 \frac 3 4 \cdots \frac{2n-2}{2n-1}\frac{2n-1}{2n}=\frac 1 {4n}$$

Thus, $\sum a_n$ diverges.

• Very clever! Thank you Jul 3, 2013 at 23:27

Since $$\prod_{k=1}^n\frac{2k-1}{2k}\times\prod_{k=1}^n\frac{2k}{2k+1}=\frac1{2n+1}$$ and \begin{align} 1 &\ge\left.\prod_{k=1}^n\frac{2k-1}{2k}\middle/\prod_{k=1}^n\frac{2k}{2k+1}\right.\\ &=\prod_{k=1}^n\frac{4k^2-1}{4k^2}\\ &\ge\prod_{k=1}^\infty\frac{4k^2-1}{4k^2}\\ &=\prod_{k=1}^\infty\left(1+\frac1{4k^2-1}\right)^{-1}\\ &\ge\exp\left(-\sum_{k=1}^\infty\frac1{4k^2-1}\right)\\ &=\exp\left(-\frac12\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\right)\\ &=e^{-1/2} \end{align} we have that $$\frac{e^{-1/2}}{2n+1}\le\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2\le\frac1{2n+1}$$ Thus, since $a_n\ge\dfrac{e^{-1/2}}{2n+1}$, the series diverges.

I just read that there is a partial converse to Raabe's test that if $\frac{a_{n+1}}{a_n}\geq 1-p/n$ for $p\leq 1$, then the series diverges. So I think that settles that it diverges.