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). $$\dfrac{1}{\sqrt{n+1}}>\prod_{k=1}^n\dfrac{2k}{2k+1}>\dfrac{1}{n+1}$$