Show that $\lim_{m \to \infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0$

Question

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:

1. Proving the mentioned limits.

2. 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 $$\dfrac{(2m+1)!(2m+3)}{2^{2m+1}[(m+1)!]^2} = \dfrac{\prod_{k=1}^{2m+3}k}{(2m+2)2^{2m+1}\prod_{k=1}^{m+1}k^2} =\dfrac{\prod_{k=1}^{2m+3}k}{(m+1) \prod_{k=1}^{m+1}(2k)^2} = \dfrac{2m+3}{m+1} \dfrac{1 \times 3 \times \dots \times (2m+1)}{2 \times 4 \times \dots \times (2m+2)} \ge \dfrac{2m+3}{m+1} \dfrac{1}{\prod_{k=1}^{2m+1}(1+1/k)} \to 0$$

Answer 1

\begin{align} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)} % &= \frac{\left(2^{m}\right)^2 \left(\prod_{k=1}^m k\right)^2(m+1)}{\left(\prod_{k=1}^{2m+1} k\right)(2m+1)} % = \frac{\left(\prod_{k=1}^m (2k)\right)^2(m+1)}{\left(\prod_{k=1}^{m} (2k)(2k+1)\right)(2m+1)}\\ &= \frac{1}{\prod_{k=1}^m \frac{(2k)(2k+1)}{(2k)^2}}\frac{m+1}{2m+1} = \frac{1}{\prod_{k=1}^m \left(1+\frac{1}{2k}\right)}\frac{m+1}{2m+1} \end{align}

Since you wanted to learn some infinite product stuff (as opposed to just using Sterling's formula), from wikipedia we have $$1+\sum_{n=1}^N p_n \leq \prod_{n=1}^N(1+p_n)\leq\exp\left(\sum_{n=1}^Np_n\right)$$

Applying this to the denominator product above, we can bound:

$$\frac{1}{\exp\left(\sum_{k=1}^m \frac{1}{2k}\right)}\frac{m+1}{2m+1} \leq \frac{1}{\prod_{k=1}^m \left(1+\frac{1}{2k}\right)}\frac{m+1}{2m+1} \leq \frac{1}{1 +\sum_{k=1}^m \frac{1}{2k}}\frac{m+1}{2m+1}$$

Clearly $\frac{m+1}{2m+1} \to \frac{1}{2}$, and we know $H_m=\sum_{k=1}^m\frac{1}{k} \to \infty$ as it is partial sums of the Harmonic series, hence $$\underbrace{\frac{1}{\exp\big(\frac12 \underbrace{H_m}_{\to\infty}\big)}}_{\to 0}\underbrace{\frac{m+1}{2m+1}}_{\to \frac12} \leq \frac{1}{\prod_{k=1}^m \left(1+\frac{1}{2k}\right)}\frac{m+1}{2m+1} \leq \underbrace{\frac{1}{1 +\frac12 \underbrace{H_m}_{\to\infty}}}_{\to 0}\underbrace{\frac{m+1}{2m+1}}_{\to \frac12}$$

the left and right sides of the inequality approach $0$ and it follows that $$\lim_{m\to\infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)} = 0$$ by the squeeze theorem.

Unfortunately this invalidates your second claim, note that \begin{align} \frac{(2m+1)!(2m+3)!}{2^{2m+1}[(m+1)!]^2} &= \left[\frac{2^{2m+1}[(m+1)!]^2}{(2m+1)!(2m+3)!}\right]^{-1}\\ &= \left[\frac{2(m+1)(2m+1)}{(2m+3)!}\frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}\right]^{-1}\\ &= \bigg[\frac{2(m+1)(2m+1)}{(2m+1)!(2m+2)(2m+3)}\frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)} \bigg]^{-1}\\ &= \bigg[\underbrace{\frac{1}{(2m+1)!}}_{\to 0}\underbrace{\frac{2m+1}{2m+3}}_{\to 1}\underbrace{\frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}}_{\to 0} \bigg]^{-1} \end{align} where I've intentionally refactored in terms of the limit that we already determined. Note that denominator goes to $0$ and strictly positive, hence the expression goes to $+\infty$, so: $$\lim_{m\to\infty} \frac{(2m+1)!(2m+3)!}{2^{2m+1}[(m+1)!]^2} = \infty$$ If you made a mistake when you derived that expression then you should now have the skills to prove the hypothetically correct expression.

As far as books go, I'd recommend Knopp's Theory and Application of Infinite Series. It has a pretty decent chapter showing how you can apply infinite series techniques to infinite products.

Added In response to your edit, you have actually only shown the lower bound. Indeed, all terms are non-zero, so $0$ is a trivial lower bound anyway.

We first derive a simple upperbound on $H_m$: $$ H_m = \sum_{k=1}^m \frac{1}{k} = 1 + \sum_{k=2}^m \frac{1}{k} \underbrace{\int_{k-1}^k \;dx}_{=\;1} \leq 1 + \sum_{k=2}^m \int_{k-1}^k \frac{dx}{x} =1+\int_1^m\frac{dx}{x} = 1 + \log(m) $$ hence \begin{align} 0 &\leq \frac{(2m+1)!(2m+3)}{2^{2m+1}(m+1)!^2} = \frac{\left(\prod_{k=1}^{2m+1}k\right) (2m+3)}{2\cdot (2^m)^2 \left(\prod_{k=1}^m k\right)^2(m+1)^2}\\ &= \frac{\left(\prod_{k=1}^{m}(2k)(2k+1)\right) (2m+3)}{2\cdot \left(\prod_{k=1}^m(2k)\right)^2(m+1)^2} = \prod_{k=1}^m \left(1+\frac{1}{2k}\right) \cdot \frac{2m+3}{2(m+1)^2}\\ &\leq \exp(\tfrac12 H_m) \frac{2m+3}{2(m+1)^2}\\ &\leq \exp(\tfrac12 (1+\log(m))) \frac{2m+3}{2(m+1)^2} = \frac{\sqrt{em}(2m+3)}{(m+1)^2} \to 0 \end{align}

Answer 2

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.

Answer 3

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 $$\lim_{m \to \infty} \dfrac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=\lim_{m \to \infty}\sqrt{\dfrac{2\pi \ m(m+1)^3}{(2m+1)^5}}\to0$$

Answer 4

$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} $

Answer 5

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 $\frac{12929}{51200}\sqrt{\frac{\pi }{10}}\approx 0.141537$