Limit as $n\to+\infty$ of $\prod_{k=1}^{n} \frac{2k}{2k+1}$ I'm trying to evaluate
$$\lim_{n\to+\infty} \prod_{k=1}^{n} \frac{2k}{2k+1}$$
First I notice that since $k\geq1$ it is $\frac{2k}{2k+1}>0$ for all $k\in\{1,...,n\}$; so
$$0\leq\lim_{n\to+\infty} \prod_{k=1}^{n} \frac{2k}{2k+1}$$
Then I notice that
$$\prod_{k=1}^{n} \frac{2k}{2k+1}=\exp{\ln\left(\prod_{k=1}^{n} \frac{2k}{2k+1}\right)}=\exp{\sum_{k=1}^{n}\ln\left(\frac{2k}{2k+1}\right)}=$$
$$=\exp{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k+1}\right)}$$
Since $\ln(1+x)\leq x$ for all $x>-1$ and since $\exp$ is an increasing function it follows that
$$\exp{\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k+1}\right)}\leq\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}$$
So
$$\lim_{n\to+\infty}\prod_{k=1}^{n} \frac{2k}{2k+1}\leq\lim_{n\to+\infty}\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}$$
Since $\exp$ is a continuous function it follows that
$$\lim_{n\to+\infty}\exp{\sum_{k=1}^{n}-\frac{1}{2k+1}}=\exp{\sum_{k=1}^{+\infty}-\frac{1}{2k+1}}=e^{-\infty}=0$$
So by the comparison test we deduce that the limit is $0$.
Is this correct? Thanks for your time.
 A: Your caclulations are correct. 
But I thought it might be interesting to see another nice trick.


*

*Let $A_n = \prod_{k=1}^n\frac{2k}{2k+1}$ and $B_n = \prod_{k=1}^n\frac{2k+1}{2k+2}$.


Then, $A_n < B_n$ and $A_nB_n$ is a telescoping product and you get
$$A_n^2 < A_nB_n = \frac{2}{2n+2}=\frac 1{n+1}$$
Hence,
$$0 < A_n < \frac{1}{\sqrt{n+1}}\stackrel{n\to \infty}{\longrightarrow}0$$
A: Yours is fine. Another simple method would be
$$\lim_{n\to\infty}\prod_{k=1}^n\dfrac{2k}{2k+1}<\lim_{n\to\infty}\prod_{k=1}^n\dfrac{2k}{2k+2}\\ = \lim_{n\to\infty}\dfrac{1}{n+1}\\=0$$
EDIT
As Dr.WolfgangHintze pointed out in the comments, this inequality is in the reverse direction, so now we have the rate of decay of this series (from other answers).
$$\dfrac{1}{\sqrt{n+1}}>\prod_{k=1}^n\dfrac{2k}{2k+1}>\dfrac{1}{n+1}$$
A: Consider
$$a_n=\prod_{k=1}^{n} \frac{2k}{2k+1}\implies \log(a_n)=\sum_{k=1}^{n}\log \left(\frac{2 k}{2 k+1}\right)=-\sum_{k=1}^{n}\log \left(1+\frac{1}{2 k}\right) $$ Use Taylor expansion
$$\log \left(1+\frac{1}{2 k}\right)=\frac{1}{2 k}-\frac{1}{8 k^2}+O\left(\frac{1}{k^3}\right)$$ Compute the sum for this truncated series to get
$$\log(a_n)\sim-\frac{H_n}{2}+\frac{H_n^{(2)}}{8}$$ where appear generalized harmonic numbers.
Use the asymptotics for large $n$ to get
$$\log(a_n) =\left(\frac{\pi
   ^2}{48}-\frac{\gamma }{2}\right)-\frac{1}{2} \log \left({n}\right)-\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$
$$a_n\sim \frac{ \exp\left(\frac{\pi
   ^2}{48}-\frac{\gamma }{2}\right)}{\sqrt n}$$
A: Another way:
Using arithmetic geometric Inequality
$$\frac{k+k-1}{2}> \sqrt{k\cdot(k-1)}\Rightarrow \frac{2k-1}{2k}>\sqrt{\frac{k-1}{k}}$$
$$\frac{2k}{2k-1}<\sqrt{\frac{k-1}{k}}\Rightarrow \prod^{n+1}_{k=2}\frac{2k}{2k-1}<\prod^{n+1}_{k=2}\sqrt{\frac{k-1}{k}}=.\frac{1}{\sqrt{n+1}}$$
$$\Longrightarrow 0<\prod^{n+1}_{k=2}\frac{2k}{2k-1}<\frac{1}{\sqrt{n+1}}$$
Applying limit $n\rightarrow \infty$ and Using Squeeze Theorem 
We have $$\prod^{n+1}_{k=2}\frac{2k}{2k-1}=0$$
