Determine when $\sum_1^\infty \frac{(2n)!x^n}{n(n!)^2}$ converges. Determine when $\sum_1^\infty \frac{(2n)!x^n}{n(n!)^2}$ converges.
By ratio test, when $|x|<1/4$, the sum converges, $|x|>1/4$ diverges. But I'm not sure about $|x|=1/4$.
By Stirling's approximation $\frac{(2n)!(1/4)^n}{n(n!)^2}\sim\frac{4^n}{n{4^n}\sqrt{ \pi n}}\sim\frac{1}{\sqrt{\pi}n^{3/2}}$, so when $|x|=1/4$, the sum converges.
But I haven't learned Stirling's approximation yet, so how to prove that when $|x|=1/4,$ the sum converges, without using it?

Let's study what happens when $x=1/4$.
Define $a_n:=\frac{(2n)!x^n}{n(n!)^2} $, $b_n:=\dfrac{1}{n^{5/4}}$ and observe the following growth rates:
$$ \dfrac{a_{n+1}}{a_n} 
=  \frac{(2n+2)!x^{n+1}}{(n+1)((n+1)!)^2} \frac{n(n!)^2}{(2n)!x^n} 
= \dfrac{(2n+2)(2n+1)n}{4(n+1)^3} 
= \dfrac{(2n+1)n}{2(n+1)^2}=\big(1-\dfrac{1}{n+1}\big) \big(1-\dfrac{1}{2(n+1)}\big)$$
$$ \dfrac{b_{n+1}}{b_n} =\dfrac{n^{5/4}}{(n+1)^{5/4}} = \big(1-\dfrac{1}{n+1}\big) \big(1-\dfrac{1}{n+1}\big)^{1/4}$$
As $n\to +\infty$
$$\dfrac{b_{n+1}}{b_n} -\dfrac{a_{n+1}}{a_n}
=\big(1-\dfrac{1}{n+1}\big)
\bigg[\big(1-\dfrac{1}{n+1}\big)^{1/4} - \big(1-\dfrac{1}{2(n+1)}\big) \bigg]
=\big(1-\dfrac{1}{n+1}\big)
\bigg[\dfrac{1}{4(n+1)} + o\big(\dfrac{1}{n+1}\big) \bigg]=\big(1-\dfrac{1}{n+1}\big) \dfrac{1+o(1)}{4(n+1)}$$
The expression $1+o(1)$ is evenually positive, therefore the RHS is eventually positive and so there exists $N > 0$ s.t.:
$$a_{n+1}/a_n \leq b_{n+1}/b_{n} \qquad \forall n\geq N \qquad (1)$$
For any integer $k > N$, considering repeatedly $(1)$ for $ N \leq n\leq k-1$ and multiplying termwise those inequalities we get:
$$a_{k}/a_N \leq b_{k}/b_{N}$$
This means:
$$a_{k}\leq b_{k}\frac{a_N}{b_{N}} \qquad \forall k \geq N$$
Therefore
$$
\sum_{k=N}^{+\infty} a_k \leq 
\sum_{k=N}^{+\infty} b_{k}\frac{a_N}{b_{N}} \leq \ +\infty$$
showing that $\sum a_n$ converges.
This also proves that the series converges absolutely if $x=-1/4$.
