Consider the following series:

\begin{align*}

a_1 &= \frac{4}{3}\\

a_2 &= \frac{4}{3}\frac{9}{8}\\

&\vdots\\

a_n &= \frac{4}{3}\frac{9}{8}\cdots\frac{(n+1)^2}{(n+1)^2-1}

\end{align*}

In other words, $a_n$ is the product of all the numbers of the form $n^2/(n^2 – 1)$ for $n=2,\dots, n+1$.

Does $\lim_{n\to\infty} a_n$ exist?

Yes, it does! Because $n$ divides $(n+1)^2 – 1$, we can cancel a bunch of terms out and rewrite

$$a_n = \frac{2(n+1)^2}{(n+1)(n+2)}.$$

Thus $\lim_{n\to\infty} a_n = 2$.

But what about $a_n = \prod_{i=2}^{n+1} n^\alpha/(n^\alpha-1)$? For which $\alpha\in [1,2]$ is $\lim_{n\to \infty} a_n$ finite? It is not for $\alpha = 1$ but as we have seen, it is for $\alpha = 2$. Does this limit exist for $\alpha \gt 1$?