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.

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 $$\prod^{n+1}_{k=2}\frac{2k}{2k-1}=0$$
