Limit as $n\to+\infty$ of $\prod_{k=1}^{n} \frac{2k}{2k+1}$
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 $$\prod^{n+1}_{k=2}\frac{2k}{2k-1}=0$$
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,
$$0 < A_n < \frac{1}{\sqrt{n+1}}\stackrel{n\to \infty}{\longrightarrow}0$$
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). $$\dfrac{1}{\sqrt{n+1}}>\prod_{k=1}^n\dfrac{2k}{2k+1}>\dfrac{1}{n+1}$$
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 $$\log(a_n) =\left(\frac{\pi ^2}{48}-\frac{\gamma }{2}\right)-\frac{1}{2} \log \left({n}\right)-\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$
$$a_n\sim \frac{ \exp\left(\frac{\pi ^2}{48}-\frac{\gamma }{2}\right)}{\sqrt n}$$