Find a limit of the recursive sequence The task is to prove sequence convergence and find a limit.
$x_0=0$
$x_1=1$
$x_{n+1}=\frac {x_n + n \cdot x_{n-1}} {n+1}$
I have computed some values of a sequence to build up some idea of the data:
elements with even indexes converge from 0 to ~0.68, and elements with odd indexes converge from 1 to the same value.
It's obvious that the sequence isn't monotonic, hence I had to stick with a Cauchy theorem. But it led me nowhere:
$|x_{n+p}-x_{n}| < | \frac {n \cdot x_{n-1}} {n + 1} + 2 \cdot \sum_{i=n}^{2n-2} x_i + { \frac {x_{2n-1}} {2n} } |$
(I got there under the assumption that $n = p$.)
Then I tried another move:
$x_{n+1} - x_{n} = \frac {x_n + n \cdot x_{n-1}} {n+1} - x_n = $
$= \frac {-n \cdot ( x_n - x_{n-1})} {n+1}$
$x_{n} - x_{n-1} = \frac {(1-n) \cdot (x_{n-1} - x_{n-2})} {n}$
It looks like progress, but I still don't know how to go next.
 A: You're on the right track. If you write $d_n=x_{n+1}-x_n$, you have
$d_n=-\frac n{n+1}d_{n-1}$
So $d_n=(-\frac n{n+1})(-\frac {n-1}{n})(-\frac {n-2}{n-1})\ldots\frac12d_0=\frac{(-1)^n}{n+1}d_0=\frac{(-1)^n}{n+1}$
Then $x_n=x_0+\sum\limits_{k=0}^{n-1}d_k=\sum\limits_{k=0}^{n-1}d_k$
And $\lim\limits_{n\to\infty}x_n=\sum\limits_{n=0}^\infty d_n=\sum\limits_{n=0}^\infty \frac{(-1)^n}{n+1}$ which is a well known expansion of $\log 2$.
A: as you have computed $x_{n+1}-x_n=-n(x_n-x_{n-1})/(n+1)$
we have $(n+1)(x_{n+1}-x_n)=-n(x_n-x_{n-1})$
then denote $g(n) = n(x_n-x_{n-1})$
we have $g(n+1) = -g(n)$
then $g(n) = (-1)^{n-1}g(1) = (-1)^{n-1}$
so $n(x_n-x_{n-1}) = (-1)^{n-1}$
$x_n-x_{n-1} = \frac{(-1)^{n-1}}{n}$
$x_n-x_m = \sum_{i=m+1}^{n}\frac{(-1)^{i-1}}{i}$
Now convergence of $x_n$ follows from convergence of $\sum{\frac{(-1)^n}{n}}$ and $x_n$ converges to $\ln{2}$
A: The sequence converges to $\log 2$.
You can show inductively the following:
a. $x_{2n}$ is increasing,
b. $x_{2n-1}$ is decreasing,
c. $x_{n+1}-x_n=\dfrac{(-1)^n}{n+1}$. 
d. $x_n=\displaystyle\sum_{k=0}^n \dfrac{(-1)^{k-1}}{k}\to\log 2$.
