<s>[INST] Show that $\lim_{m \to \infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0$ After a long way to solve the differential equation $y''+xy'+3y=0$, I arrived at the solution $$y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}$$ and now I need to prove that it is convergent for all $x \in \mathbb{R}$ as the book claims.
Applying The Ratio Test, I will need just to show that the two following limits hold:  $$\lim_{m \to \infty} \dfrac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0 \ \ \ \text{and} \ \ \lim_{m \to \infty} \dfrac{(2m+1)!(2m+3)!}{2^{2m+1}[(m+1)!]^2}=0. $$ But after lots of effort I don't know how to prove that the limits are zero and unfortunately I know not much about infinite products as all Real Analysis books teach only infinite sums! So my questions are:


*

*Proving the mentioned limits.

*A self-learn-able book containing lots of theorems and information about infinite products.  
--
Added. [after reading @adfriedman's answer]: I have made a mistake regarding the second limit and the correct one is to show that $ \lim_{m \to \infty} \dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2}=0$ which holds since $$\dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2} = \dfrac{\prod_{k=1}^{2m+3}k}{(2m+2)2^{2m+1}\prod_{k=1}^{m+1}k^2} =\dfrac{\prod_{k=1}^{2m+3}k}{(m+1) \prod_{k=1}^{m+1}(2k)^2} = \dfrac{2m+3}{m+1} \dfrac{1 \times 3 \times  \dots \times (2m+1)}{2 \times 4 \times  \dots \times (2m+2)} \ge \dfrac{2m+3}{m+1} \dfrac{1}{\prod_{k=1}^{2m+1}(1+1/k)} \to 0$$   
 [/INST] $y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}
$
Use the root test
and Stirling
($n!\approx \sqrt{2\pi n}(n/e)^n$).
$\begin{array}\\
\dfrac{2^m(m+1)!}{(2m+1)!}
&=\dfrac{m+1}{2m+1}\dfrac{2^mm!}{(2m)!}\\
&\approx\dfrac12\dfrac{2^m\sqrt{2\pi m}(m/e)^m}{\sqrt{4\pi m}(2m/e)^{2m}}\\
&=\dfrac12\dfrac1{\sqrt{2}}\left(\dfrac{2m/e}{4m^2/e^2}\right)^m\\
&=\dfrac1{2\sqrt{2}}\left(\dfrac{e}{2m}\right)^m\\
\text{so}\\
\left(\dfrac{2^m(m+1)!}{(2m+1)!}\right)^{1/m}
&\approx\dfrac1{(2\sqrt{2})^{1/m}}\left(\dfrac{e}{2m}\right)\\
&\to 0\\
\end{array}
$
For the first one:
$\begin{array}\\
\dfrac{2m+1}{2^mm!}
&=(2m+1)\dfrac{1}{2^mm!}\\
&\approx(2m+1)\dfrac{1}{2^m\sqrt{2\pi m}(m/e)^m}\\
&=\dfrac{2m+1}{\sqrt{2\pi m}}\dfrac{1}{2^m(m/e)^m}\\
&=\dfrac{2m+1}{\sqrt{2\pi m}}\left(\dfrac{e}{2m}\right)^m\\
\text{so}\\
\left(\dfrac{2m+1}{2^mm!}\right)^{1/m}
&\approx\left(\dfrac{2m+1}{\sqrt{2\pi m}}\right)^{1/m}\left(\dfrac{e}{2m}\right)\\
&\to 0
\qquad\text{since } m^{1/m} \to 1\\
\end{array}
$
</s>