Limit of certain recurrence relation So given this recurrence relation (not how it was presented, but equivalent and much nicer)
$$ x_{n+1} = \dfrac{x_n + nx_{n-1}}{n+1}; \ x_0 = 0,\ x_1 = 1 $$
I just can't find what the limit as $n$ goes to infinity is. It was straightforward to show it was convergent, and oscillatory about its limit, and I know the limit is going to be between $0.6$ and $0.75$ (through repeated calculation). The next step of finding the exact limit, though, I'm completely lost on. 
 A: If $$S_n = \sum_{j=1}^{n} \frac{(-1)^{j+1}}{j}$$
then we can prove that
$$(n+1)S_{n+1} = S_n + nS_{n-1}$$
Since $x_1 = S_1$ and $x_2 = S_2$ we have that $x_n = S_n$ for $n \ge 1$
Thus $\lim x_n = \lim S_n = \log 2$
A: Define ${\rm F}\left(z\right) \equiv \sum_{n = 0}^{\infty}z^{n}\,x_{n}\ \ni\ {\rm F}\left(0\right) = 0$ and ${\rm F}\,'\left(1\right) = 1$. Then
\begin{eqnarray*}
0
& = &
\sum_{n = 1}^{\infty}z^{n}\left\lbrack%
\left(n + 1\right)x_{n + 1} - x_{n} - nx_{n - 1}
\right\rbrack
\\
& = &
\sum_{n = 2}^{\infty}z^{n - 1}\,n\,x_{n}
-
\sum_{n = 1}^{\infty}z^{n}\,x_{n} - \sum_{n = 0}^{\infty}z^{n + 1}\left(n + 1\right)x_{n}
\\
& = &
\left\lbrack{\rm F}\,'\left(z\right) - 1\right\rbrack - {\rm F}\left(z\right) - \left\lbrack z^{2}{\rm F}\,'\left(z\right)
+
z{\rm F}\left(z\right)\right\rbrack
\\
& = &
\left(1 - z^{2}\right){\rm F}\,'\left(z\right) - \left(1 + z\right){\rm F}\left(z\right) - 1
\end{eqnarray*}
$$
\left(z - 1\right){\rm F}\,'\left(z\right)
+
{\rm F}\left(z\right)
=
-\,{1 \over z + 1}
\quad\Longrightarrow\quad
{{\rm d}\left\lbrack\left(z - 1\right){\rm F}\left(z\right)\right\rbrack \over {\rm d}z}
=
-\,{1 \over z + 1}
$$
$$
\left(z - 1\right){\rm F}\left(z\right)
=
-\int_{0}^{z}{{\rm d}z \over z + 1}
=
-\ln\left(1 + z\right)
$$
$$
\quad\Longrightarrow\quad
{\rm F}\left(z\right)
=
{\ln\left(1 + z\right) \over 1 - z}
=
z+z^2/2+(5 z^3)/6+(7 z^4)/12+(47 z^5)/60+O(z^6)
$$
$$
x_{0} = 0,\ x_{1} = 1,\ x_{2} = {1 \over 2},\ x_{3} = {5 \over 6}\,\ x_{4} = {7 \over 12}\,\ x_{5} = {47 \over 60}\,\quad\mbox{etc...}
$$
In general
\begin{align}
{\rm F}\left(z\right)
&=
{\ln\left(1 + z\right) \over 1 - z}
=
\sum_{s = 0}^{\infty}z^{s}\sum_{k = 0}^{\infty}{\left(-1\right)^{k} \over k + 1}\,z^{k + 1}
\sum_{n = 1}^{\infty}\delta_{n, s + k + 1}
=
\sum_{n = 1}^{\infty}z^{n}
\sum_{s = 0}^{\infty}\sum_{k = 0}^{\infty}{\left(-1\right)^{k} \over k + 1}\,
\delta_{k,n - s -1}
\\[3mm]&=
\sum_{n = 1}^{\infty}z^{n}
\sum_{s = 0}^{n - 1}{\left(-1\right)^{n - s -1} \over \left(n - s -1\right) + 1}
=
\sum_{n = 1}^{\infty}z^{n}
\sum_{s = 0}^{n - 1}{\left(-1\right)^{s} \over s + 1}
\end{align}
$$
\quad\Longrightarrow\quad
x_{n} = \sum_{s = 0}^{n - 1}{\left(-1\right)^{s} \over s + 1}\,,
\qquad
n \geq 1
$$
$$
\lim_{n \to \infty}x_{n}
=
\sum_{s = 0}^{\infty}{\left(-1\right)^{s} \over s + 1}
=
\sum_{s = 0}^{\infty}{\left(-1\right)^{s} \over s + 1}\ 1^{s + 1}
=
\ln\left(1 + 1\right)
=
{\Large \ln\left(2\right)}
$$
