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 $$\dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2} = \dfrac{\prod_{k=1}^{2m+3}k}{(2m+2)2^{2m+1}\prod_{k=1}^{m+1}k^2} =\dfrac{\prod_{k=1}^{2m+3}k}{(m+1) \prod_{k=1}^{m+1}(2k)^2} = \dfrac{2m+3}{m+1} \dfrac{1 \times 3 \times  \dots \times (2m+1)}{2 \times 4 \times  \dots \times (2m+2)} \ge \dfrac{2m+3}{m+1} \dfrac{1}{\prod_{k=1}^{2m+1}(1+1/k)} \to 0$$   
 A: $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}
$
A: Let's attempt to expand the factorials in the first limit and hopefully cancel some terms. Hopefully, then a similar approach will help on the second limit.
$$\lim_\limits{m \to \infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)} = \lim_\limits{m \to \infty} \frac{2^{2m}(1\cdot2\cdots m)(1\cdot2\cdots m \cdot (m+1))}{(1 \cdot 2 \cdot 2m)(2m+1)^2}.$$
Since we are multiplying the numerator by $2$ $2m$ times, we can distribute the $2^{2m}$ as follows:
$$\lim_\limits{m \to \infty} \frac{(2\cdot4\cdots 2m)(2\cdot4\cdots 2m) (m+1)}{(1 \cdot 2 \cdot 2m)(2m+1)^2} = \lim_\limits{m \to \infty} \frac{(2\cdot4\cdots 2m) (m+1)}{(1 \cdot 3 \cdot 5 \cdots (2m - 1))(2m+1)^2}$$
$$= \lim_\limits{m \to \infty} \frac{(2\cdot4\cdots 2m) (m+1)}{(3 \cdot 5 \cdots (2m + 1))(2m+1)} = \lim_\limits{m \to \infty} \left( \prod_{1 \leq i \leq m} \frac{2i}{2i+1} \right) \frac{m + 1}{2m + 1}$$
$$= \lim_\limits{m \to \infty} \prod_{1 \leq i \leq m} \frac{2i}{2i+1} \lim_\limits{m \to \infty} \frac{m + 1}{2m + 1}.$$
As you pointed out, $\lim_{m \to \infty} (m + 1)/(2m + 1) = 1/2$, so we'll have to show instead that $\lim_\limits{m \to \infty} \prod_{1 \leq i \leq m} \frac{2i}{2i+1} = 0$ and then we'll be done. A theorem from Wikipedia says that an infinite product $\prod_{1 \leq i \leq \infty} a_n$ converges to a nonzero real number if and only if the sum $\sum_{1 \leq i \leq \infty} \log(a_n)$ converges. So, let's try to show that the sum $\sum_{1 \leq i \leq \infty} \log(a_n) = \sum_{1 \leq i \leq \infty} \log((2i)/(2i + 1))$ diverges, and then we can conclude that the infinite product, and the original limit, are $0$.
One approach could be the limit comparison test. This test only works on positive series, so we'll test the convergence of
$$-\sum_{1 \leq i \leq \infty} \log((2i)/(2i + 1)) = \sum_{1 \leq i \leq \infty} \log((2i + 1)/(2i)) = \sum_{1 \leq i \leq \infty} b_n.$$
The convergence of the negative form of a series implies the convergence of the positive form, and likewise for divergence.
Try comparing this sum with the harmonic series, the sum of $1/i$:
$$\lim\limits_{m \to \infty} \frac{b_m}{1/m} = \frac{1}{\ln 10} \lim\limits_{m \to \infty} \frac{\ln((2m + 1)/(2m))}{1/m} = \frac{1}{\ln 10}\lim\limits_{m \to \infty} \frac{\frac{2m}{2m + 1} \cdot \frac{2m(2) - (2m + 1)2}{(2m)^2}}{-1/m^2} \text{ (L'Hospitals)} = \frac{1}{\ln 10}\lim\limits_{m \to \infty} -m^2 \frac{2m}{2m + 1} \cdot \frac{4m - (4m + 2)}{(2m)^2} = \frac{1}{\ln 10}\lim\limits_{m \to \infty} \frac{-2m}{2m + 1} \cdot \frac{-2}{4} = \frac{1}{2\ln 10}\lim\limits_{m \to \infty} \frac{2m}{2m + 1} = \frac{1}{2\ln 10} > 0.$$
Since the above limit is positive and nonzero, $\sum_{1 \leq i \leq \infty} b_n$ has the same convergence as the divergent harmonic series $\sum_{1 \leq i \leq \infty} (1/i)$. So $\sum_{1 \leq i \leq \infty} b_n = \sum_{1 \leq i \leq \infty} \log(a_n)$ is divergent, and from the theorem from Wikipedia above, then $\prod_{1 \leq i \leq \infty} = 0$.
As far as the second question about a book, I'm not sure as I haven't read any particularly outstanding books on the topic. But the methods I used here are taught from calculus 2. Consequently, this is a more basic approach; other approaches might use higher-level math.
A: With Stirling's formula approximation
$$n!\sim \sqrt{2\pi n}\left(\frac{n^n}{e^n}\right)$$ as $n\to \infty$
for the first one
$$\lim_{m \to \infty} \dfrac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=\lim_{m \to \infty}\sqrt{\dfrac{2\pi \ m(m+1)^3}{(2m+1)^5}}\to0$$
A: If you want to go beyond and even be able to approximate the value, consider $$A_m=\frac{2^{2m}\,m!\,(m+1)!}{(2m+1)!\,(2m+1)}$$
Take logarithms
$$\log(A_m)=2m \log(2)+\log(m!)+\log((m+1)!)-\log((2m+1)!)-\log(2m+1)$$ Use Stirling approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ Apply it to each fcatorial and continue with Taylor expansion for large values of $m$. You should arrive to
$$\log(A_m)=-\frac{1}{2} \log \left({m}\right)+\log \left(\frac{\sqrt{\pi
   }}{4}\right)+\frac{1}{8 m}-\frac{1}{4 m^2}+O\left(\frac{1}{m^3}\right)$$ Continuing with Taylor series
$$A_m=e^{\log(A_m)}=\frac 14 \sqrt{\frac \pi m}\left(1+\frac{1}{8 m}-\frac{31}{128 m^2}+O\left(\frac{1}{m^3}\right)\right)$$
For $m=10$, the exact value is $\frac{262144}{1851759}\approx 0.141565$ while the above approximation would give $\frac{12929}{51200}\sqrt{\frac{\pi }{10}}\approx 0.141537$
