Prove that $(y_n)$ has a finite limit when $n \rightarrow \infty$. Let a real sequence $(a_n)$ defined by:
$x_1$ = 1; $x_n = \displaystyle\frac{2n}{(n-1)^2} \displaystyle\sum_{i=1}^{n-1} x_i$, $n=2,3,4, \dots$
For each integers $n$ let
$y_n = x_{n+1} - x_n$
Prove that $(y_n)$ has a finite limit when $n \rightarrow \infty$.

My attempt: We have

$x_{n+1} = \displaystyle\frac{2(n+1)}{n^2} \displaystyle\sum_{i=1}^{n} x_i = \displaystyle\frac{2(n+1)}{n^2}\left(\frac{2n}{(n-1)^2}+1\right) \left(\displaystyle\frac{2(n-1)}{(n-2)^2}+1\right) \dots 5 x_1$.
Hence, when $n \rightarrow \infty$, $(x_n)$ has a finte limit $\Rightarrow$ $(y_n)$ has a finite limit.
Is that right? Thank all!

Not really an answer, just a clarification: $(x_n)_n$ indeed does not converge.
We have:
\begin{align}
x_{n+1} &= \frac{2(n+1)}{n^2}\left(\frac{2n}{(n-1)^2} + 1\right)\left(\frac{2(n-1)}{(n-2)^2} + 1\right)\cdots\left(\frac{2\cdot 2}{1^2} + 1\right)\\
&= \frac{2(n+1)}{n^2}\cdot\frac{2n + (n-1)^2}{(n-1)^2}\frac{2(n-1) + (n-2)^2}{(n-2)^2}\cdots\frac{2\cdot 2 + 1^2}{1^2}\\
&= \frac{2(n+1)}{n^2}\cdot\frac{n^2+1}{(n-1)^2}\frac{(n-1)^2 + 1}{(n-2)^2}\cdots\frac{2^2+1}{1}\\
&= \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2+1}{(k-1)^2}\\
&\ge \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2}{(k-1)^2}\\
&\ge \frac{2(n+1)}{n^2}\cdot\left(\frac{n!}{(n-1)!}\right)^2\\
&= \frac{2(n+1)}{n^2} \cdot n^2\\
&= 2(n+1)
\end{align}
so $(x_n)_n$ is unbounded.
However,
\begin{align}
y_n &= x_{n+1}-x_n\\
 &= \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2+1}{(k-1)^2} - \frac{2n}{(n-1)^2}\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
&= \left(\frac{2(n+1)}{n^2}\frac{n^2+1}{(n-1)^2} - \frac{2n}{(n-1)^2}\right)\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
&= \frac2{(n-1)^2}\left(1 + \frac1n + \frac1{n^2}\right)\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
\end{align}
which seems to converge to around $3.67608$, but it is difficult to bound the above product. Namely
$$\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2} \le \prod_{k=2}^{n-1} \frac{k^2+k}{(k-1)^2} = \prod_{k=2}^{n-1} \frac{k(k+1)}{(k-1)^2} = \frac12 \frac{(n-1)!n!}{(n-2)!} = \frac12 (n-1)^2n$$
not being good enough.
