Find limit recursion of sequence $x_{n+1} = \frac{x_n+ n x_{n-1}}{n+1} $

Question

Prove sequence

$$x_{n+1} = \frac{x_n+ n x_{n-1}}{n+1} $$

$$x_0 = 0, x_1 = 1 $$

converges and find it's limit

My attempt

1. Let's prove $0 \le x_n \le 1$:

   $x_n \ge 0 $ (obvious)

   By induction

   if $x_n \le 1$ and $x_{n-1} \le 1$ then: $$ \frac{x_n+ n x_{n-1}}{n+1} \le \frac{1+ n}{n+1}=1 \implies x_{n+1}<=1$$

2. Let's prove convergence

$$\lim_{n \to \infty }{x_{n+1}} = \lim_{n \to \infty }{ \frac{x_n + n x_{n-1}}{n+1} } = \lim_{n \to \infty }{ (\frac{x_n}{n+1} } + \frac{x_{n-1}}{1+\frac{1}{n}}) = x_{n-1}$$

So, sequence converges.

Question: I'm right so far and how to find the limit?

Thanks

Answer 1

Hint:

Check that $\displaystyle \forall n\geq 0, x_{n+1}-x_n=\frac{(-1)^n}{n+1}$

The series $\displaystyle \sum (x_{n+1}-x_n)$ is therefore convergent, and so is the sequence $(x_n)$, say $x_n\to l$

Furthermore, $\displaystyle l= \sum_{k=0}^\infty (x_{k+1}-x_k) =\sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \log 2$.

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Edit: from a simulation on Mathematica, here is the plot of the first few terms of the sequence, giving intuition about its behavior: