Prove the convergence of the recurrent sequence

Question

The sequence is given by the recurring relation:

$$ x_{0} = 0, x_{1} = 1, x_{n+1} = \frac{x_{n} + n x_{n - 1}}{n + 1} $$

Prove that the sequence converges and find its limit.

I managed to get the following, but I do not know how to prove convergence

$$ x_{n+1} - x_{n} = \frac{(-1)^n}{n+1}\\ x_{n+1} = x_{n} + \frac{(-1)^n}{n+1} = x_{n - 1} + \frac{(-1)^n}{n+1} + \frac{(-1)^{n-1}}{n} = ... = 1 + \sum_{k=1}^ n \frac{(-1)^k}{k+1} = \sum_{k=0}^ n \frac{(-1)^k}{k+1} $$

Answer 1

Since $x_0=0$ and $x_{n+1}-x_n=\dfrac{(-1)^n}{n+1}$, then$$x_{n+1}=\sum_{k=0}^nx_{k+1}-x_k=\sum_{k=0}^n\frac{(-1)^k}{k+1}$$and therefore$$\lim_{n\to\infty}x_n=\sum_{k=0}^\infty\frac{(-1)^k}{k+1}=\log 2.$$

Answer 2

$$ x_{n+1} = x_{n} + \frac{(-1)^n}{n+1} = x_{n - 1} + \frac{(-1)^n}{n+1} + \frac{(-1)^{n-1}}{n} = ... = 1 + \sum_{k=1}^{\color{red}n} \frac{(-1)^k}{k+1} = \sum_{k=0}^{\color{red}n} \frac{(-1)^k}{k+1} $$

The sign alternates and also $\frac1{k+1}$ is a positive decreasing function that congerges to zero, hence by alternating series test, $x_n$ converges.