I was given the following question :
Determine the radius of convergence of the following series : $\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$
I was able to determine using the Cauchy–Hadamard formula that $R = 4$, but I still need to determine whether the series converges for $x = 4$ or $x=-4$.
I tried using the ratio test (for $x = 4$) but the result is $1$, so I can't really do anything with it. Any ideas on how I could approach this? Thanks!
This is perhaps overkill but I think the maths is cute and it doesn't require any appeals to Stirling's approximation etc:
For $x = 4$,
$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}2^{2n}= \displaystyle\sum\limits_{n=1}^{\infty} \frac{(2n!!)^2}{(2n)!}$
$ \displaystyle\frac{(2n!!)^2}{(2n)!} = \frac{((2n)(2n-2)(2n-4)...(2))^2}{(2n)(2n-1)(2n-2)...(2)} = \frac{(2n)(2n-2)(2n-4)...(2)}{(2n-1)(2n-3)(2n-5)...(1)} = \prod_{i=1}^n{\frac{2i}{2i-1}}$
Now observe that $\displaystyle\prod_{i=1}^n{\frac{2i}{2i-1}} \ge 1$ $\forall$ $n$
And hence $\frac{(2n!!)^2}{(2n)!} \ge 1$ $\forall$ $n$
$\displaystyle\sum\limits_{n=1}^{k} \frac{(n!)^2}{(2n)!}2^{2n} \ge k $ and so certainly does not converge
For $x = -4$, notice that the terms in the sum are exactly the same as before but with alternating signs. Furthermore:
$m > n \implies \displaystyle\prod_{i=1}^m{\frac{2i}{2i-1}} > \displaystyle\prod_{i=1}^n{\frac{2i}{2i-1}}$
Hence the terms of the series are increasing in absolute value and therefore do not converge to $0$. Thus the series does not converge (this reasoning equally applies to $x=4$).
Stirling's approximation (good for intuition, even if it's not acceptable for an explanation) gives you $$ \frac{4^n(n!)^2}{(2n)!} \approx \frac{4^n(\sqrt{2\pi n} n^n e^{-n})^2} {\sqrt{2\pi(2n)} (2n)^{2n} e^{-2n}} = \frac{4^n(2\pi n) n^{2n} e^{-2n}}{2\sqrt{\pi n} 2^{2n} n^{2n} e^{-2n}} = \sqrt{\pi n} $$ Which would tell you the terms diverge individually and the series can't converge. Wolfram Alpha seems to agree.
It looks like the terms increase, in fact, and you can show this with the ratio of consecutive terms: $$ \frac{\frac{4^{n+1}((n+1)!)^2}{(2(n+1))!}}{\frac{4^n(n!)^2}{(2n)!}} = \frac{4(n+1)^2}{(2n+2)(2n+1)} = \frac{2n+2}{2n+1} > 1 $$
We have $$4^n=(1+1)^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence for $ x=\pm 4$ the absolute value of the general term is greater than $1,$ which excludes the convergence.
Remark 1 The radius of convergence is equal $4$ and can be easily determined by the $n$th root test basing on the inequality $${4^n\over 2n+1}< {2n\choose n }<4^n\quad (*)$$ as the $n$-th term in the Newton triangle is the greatest in the $2n$-th row.
Remark 2 Another series $\sum_{n=1}^\infty {(2n)!\over (n!)^2}x^n$ is more interesting. By $(*)$ the radius of convergence is equal $1/4.$ For $x=1/4$ the series is divergent due to the first inequality in $(*)$ which gives the lower bound by $(2n+1)^{-1}.$ For $x=-1/4$ the series is convergent by the Leibniz test as the sequence ${(2n)!\over 4^n(n!)^2}$ is decreasing and convergent to $0.$
