## Problem 1

Let the sequence $\left(x_{n}\right)_{n \geq 2}$ be such that $x_{2}=1, x_{n+1}=\frac{n^{2}}{n-1} \cdot x_{n}, n \geq 2$.

If $S_{n}=\sum_{k=2}^{n}\left(1-\frac{1}{k}\right) \cdot \frac{1}{x_{k}}$, calculate $\lim _{n \rightarrow \infty} S_{n}$.
## Problem 1

Let the sequence $\left(x_{n}\right)_{n \geq 2}$ be such that $x_{2}=1, x_{n+1}=\frac{n^{2}}{n-1} \cdot x_{n}, n \geq 2$. If $S_{n}=\sum_{k=2}^{n}\left(1-\frac{1}{k}\right) \cdot \frac{1}{x_{k}}$, calculate $\lim _{n \rightarrow \infty} S_{n}$.

## Solution

It is demonstrated that the sequence $\left(x_{n}\right)_{n \geq 2}$ is positive.

We have: $\quad x_{2}=1$

$$
x_{3}=\frac{2^{2}}{1} \cdot x_{2}
$$

$$
x_{n+1}=\frac{n^{2}}{n-1} \cdot x_{n}
$$

By multiplying the relations, we obtain: $x_{n+1}=\frac{(n!)^{2}}{(n-1)!}$.

Thus: $x_{n+1}=n \cdot n!, \forall n \geq 2$.

Then $S_{n}=\sum_{k=2}^{n}\left(1-\frac{1}{k}\right) \cdot \frac{1}{x_{k}}=\sum_{k=2}^{n} \frac{k-1}{k} \cdot \frac{1}{(k-1)(k-1)!}=\sum_{k=2}^{n} \frac{1}{k!}=$

$$
=\sum_{k=0}^{n} \frac{1}{k!}-2 .
$$

$\lim _{n \rightarrow \infty} S_{n}=e-2$.