how to show $a_{n}=[\frac{(2n)!!}{(2n-1)!!}]^2 \frac{1}{2n+1}$ converges?

Question

Question: $\displaystyle{a_{n} = \left[{\left(2n\right)!! \over \left(2n - 1\right)!!}\,\right]^{2} {1 \over 2n + 1}\,,\quad\mbox{prove}\ a_{n}}$ converges.

My thought: I want to prove {$a_{n}$} is an increasing sequence and it has an upper bound. I've figured out $$a_{n} = (\frac{2\cdot 2 }{1\cdot 3})\cdot(\frac{4\cdot4}{3\cdot5})\dots (\frac{(2n-2)(2n-2)}{(2n-3)(2n-1)})(\frac{(2n)^2}{2n-1})(\frac{1}{2n+1})$$ And, $$\frac{a_{n+1}}{a_{n}}=\frac{(2n)^2}{(2n-1)(2n+1)}\frac{(2n+2)^2}{(2n+1)}\frac{(2n-1)}{(2n)^2}(\frac{2n+1}{2n+3}) = \frac{(2n+2)^2}{(2n+1)(2n+3)} \gt 1$$ So it is increasing.

My problem: I'm stuck at proving it has a upper bound. Maybe there are some inequalities that can be used here?

In addition, I've searched this question and find this sequence converges to $\pi/2$ and it is known as Wallis Formula.But here, I only want to show it converges and I'm not going to find its limit.

Thanks for your time!

Answer 1

Consider the sequence $(b_n)$ defined by

$$b_n = \frac{2n+2}{2n+1}\cdot a_n.$$

Then $(b_n)$ is decreasing

\begin{align} \frac{b_{n+1}}{b_n} &= \frac{\frac{2n+4}{2n+3}}{\frac{2n+2}{2n+1}}\cdot\frac{a_{n+1}}{a_n} \\ &= \frac{(2n+4)(2n+1)}{(2n+2)(2n+3)}\cdot\frac{(2n+2)^2}{(2n+1)(2n+3)} \\ &= \frac{(2n+4)(2n+2)}{(2n+3)^2} \\ &= \frac{(2n+3)^2-1}{(2n+3)^2} \\ &< 1 \end{align}

and since $a_n > 0$ and $\frac{2n+2}{2n+1} > 1$ for all $n\geqslant 1$, we have

$$a_1 \leqslant a_n < b_n \leqslant b_1$$

for all $n \geqslant 1$, showing the boundedness of $(a_n)$.

Answer 2

Consinder $a_n = a_1 \prod \limits_{i=2}^n{\frac{a_i}{a_{i-1}}}$. You calculated that quotient above correctly: $\frac{a_i}{a_{i-1}} = \frac{(2i)^2}{(2i-1)(2i+1)} = 1+\frac{1}{(2i-1)(2i+1)}$.

The product above is thus of the general form $\prod \limits_{i=1}^{n} (1+\alpha_i)$, with $\alpha_i \ge 0$. The application of the inequality of the arithmetic and geometric means yields

$$[\prod \limits_{i=1}^{n} (1+\alpha_i)]^\frac{1}{n} \le \frac{\sum \limits_{i=1}^{n} (1+\alpha_i)}{n} = 1 + \frac{\sum \limits_{i=1}^{n} \alpha_i}{n}$$ which leads to

$$\prod \limits_{i=1}^{n} (1+\alpha_i) \le [1 + \frac{\sum \limits_{i=1}^{n} \alpha_i}{n}]^n$$.

Further, for real $x \ge 0$ we have $(1+\frac{x}{n})^n \le e^x$, because each monom of the binomial expansion of the left side ($\frac{\choose^{n}_{i}}{n^i}x^i$) is smaller or equal to the corresponding term of the Tayler expansion on the right side ($\frac{1}{n!}x^i$), and the remaining terms on the right side are all non-negative.

So we finally get

$$\prod \limits_{i=1}^{n} (1+\alpha_i) \le e^{\sum \limits_{i=1}^{n} \alpha_i}$$

That means if the infinite sum $\sum \limits_{i=1}^{\infty} \alpha_i$ converges, which is equivalent to the finite sum $\sum \limits_{i=1}^n \alpha_i$ being bounded (because $\alpha_i \ge 0$), that will make the finite product $\prod \limits_{i=1}^{n} (1+\alpha_i)$ bounded, so the infinite product $\prod \limits_{i=1}^{\infty} (1+\alpha_i)$ converges.

Now what's left to prove is that $\sum \limits_{i=2}^{\infty} \frac{1}{(2i-1)(2i+1)}$ converges, which is easy to see because $\sum \limits_{i=2}^{n} \frac{1}{(2i-1)(2i+1)} = \sum \limits_{i=2}^{n} \frac{1}{2} (\frac{1}{2i-1} - \frac{1}{2i+1}) =\frac{1}{2} (\frac{1}{3} - \frac{1}{2n+1})$

Answer 3

$$ \begin{align} & a_m=\left(\frac{(2m)!!}{(2m-1)!!}\right)^2 \frac{1}{2m+1} = \prod_{n=1}^{m}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) \Rightarrow \\ & \log(a_m)=\sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \\ & \text{Let}\space f(x) = -\log(x) \space\colon x\gt1 \space\Rightarrow f'(x) = -\frac{1}{x} \space\Rightarrow f''(x) = \frac{1}{x^2} \space\gt0 \space\Rightarrow f(x) \space\text{convex function} \\[8mm] & \text{(1).}\space f\left(\frac{x_1+x_2}{2}\right)\le\frac{f(x_1)+f(x_2)}{2}\Rightarrow \\ & \quad -\log\left(\frac{(2n-1)+(2n+1)}{2}\right) \le \frac{-\log(2n-1)-\log(2n-1)}{2} \Rightarrow \\ & \quad -2\log(2n) \le -\log(2n-1)-\log(2n-1) \Rightarrow 0\le 2\log(2n)-\log(2n-1)-\log(2n-1) \\ & \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \ge 0 \Rightarrow \log(a_m) \ge 0 \Rightarrow \quad\color{red}{a_m \ge 1} \\[8mm] & \text{(2).}\space f(x) \ge f(y)+f'(y)\,(x-y) \Rightarrow \\ & \quad -\log(2n) \ge -\log(2n-1)-\frac{2}{2n-1} \Rightarrow 0 \ge \log(2n)-\log(2n-1)-\frac{2}{2n-1} \quad\space\& \\ & \quad -\log(2n) \ge -\log(2n+1)+\frac{2}{2n+1} \Rightarrow 0 \ge \log(2n)-\log(2n+1)+\frac{2}{2n-1} \quad\Rightarrow \\ & \quad 0 \ge 2\log(2n)-\log(2n-1)-\log(2n+1)-\frac{2}{2n-1}+\frac{2}{2n+1} \Rightarrow \\ & \quad 2\log(2n)-\log(2n-1)-\log(2n+1) \le \frac{2}{2n-1}-\frac{2}{2n+1} = \frac{4}{4n^2-1} \Rightarrow \\ & \quad \sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \le \sum_{n=1}^{m}\frac{4}{4n^2-1} \Rightarrow \\ & \quad \lim_{m\rightarrow\infty} \log(a_m) \le \sum_{n=1}^{\infty}\frac{4}{4n^2-1} = 2 \Rightarrow \quad\color{red}{\lim_{m\rightarrow\infty} a_m\le e^2} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$

Answer 4

Existence of the Limit

Since $$ \begin{align} a_n &=\left[\frac{(2n)!!}{(2n-1)!!}\right]^2\frac1{2n+1}\\ &=\frac{2^{2n}n!^2}{(2n)!}\frac{2^{2n}n!^2}{(2n+1)!}\tag{1} \end{align} $$ we get $$ \begin{align} \frac{a_n}{a_{n-1}} &=\frac{4n^2}{(2n-1)2n}\frac{4n^2}{2n(2n+1)}\\ &=\frac{4n^2}{4n^2-1}\tag{2} \end{align} $$ For $x,y\ge0$, we have $(1-x)(1-y)\ge1-(x+y)$. Therefore, by induction, we have that for $x_k\ge0$, $$ \prod_{k=1}^n(1-x_k)\ge1-\sum_{k=1}^n x_k\tag{3} $$ Thus, $$ \begin{align} \frac1{a_n} &=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\\ &\ge1-\sum_{k=1}^n\frac1{4k^2}\\ &\ge1-\sum_{k=1}^\infty\frac1{4k^2-1}\\ &=1-\frac12\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\\ &=\frac12\tag{4} \end{align} $$ Therefore, $a_n$ is increasing by $(2)$ and $a_n\le2$ by $(4)$. Thus, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}a_n\text{ exists.}} $$

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Value of the Limit $$ \begin{align} \lim_{n\to\infty}a_n &=\lim_{n\to\infty}\prod_{k=1}^n\frac{2k}{2k-1}\frac{2k}{2k+1}\tag{5}\\ &=\lim_{n\to\infty}\prod_{k=1}^n\frac{k}{k-\frac12}\frac{k}{k+\frac12}\tag{6}\\ &=\lim_{n\to\infty}\frac{\Gamma(n+1)}{\left.\Gamma\left(n+\frac12\right)\middle/\Gamma\left(\frac12\right)\right.} \frac{\Gamma(n+1)}{\left.\Gamma\left(n+\frac32\right)\middle/\Gamma\left(\frac32\right)\right.}\tag{7}\\ &=\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)\lim_{n\to\infty}\frac{\Gamma(n+1)n^{-1/2}}{\Gamma\left(n+\frac12\right)}\lim_{n\to\infty}\frac{\Gamma(n+1)n^{1/2}}{\Gamma\left(n+\frac32\right)}\tag{8}\\[6pt] &=\frac\pi2\tag{9} \end{align} $$ Explanation:
$(5)$: use $(2)$
$(6)$: divide numerators and denominators by $2$
$(7)$: write products as Gamma functions and ratios of Gamma functions
$(8)$: multiply the first fraction by $n^{-1/2}$ and the second by $n^{1/2}$
$\phantom{\text{(8): }}$the limit of the product is the product of the limits
$(9)$: $\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)=\frac\pi2$ and Gautschi's Inequality says that each limit is $1$.