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