Finding the Closed Form of a Recursion I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.
So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$
Now, if we let $b_n = \frac{n(n+1)}{2} a_n$, the recursion becomes: $$b_n = \frac{n}{2} + b_{n-1}.$$
How do I continue on from here? I feel like I have to figure out a closed form for $b_n$ and use that to compute $a_n$, but I can't seem to figure out a way to do so.
Any help would be greatly appreciated.

I think all of $a_n = \frac{1}{2}$
You can easily prove it using induction also.
$a_1 = \frac{1}{2}$
Assume $a_m = \frac{1}{2}$
To prove $a_{m+1} = \frac{1}{2}$
$a_{m+1} = \frac{1}{m+2} + \frac{m+1-1}{m+2} a_m$
$a_{m+1} = \frac{1}{m+2} + \frac{m}{m+2} \frac{1}{2}$
$a_{m+1} = \frac{1}{2}$
Since the relation is true for $n=1$ and assuming the result for $a_m$, we can prove it for $a_{m+1}$, hence the result is true by induction.
Also we can get the same from your equation,
$b_n = \frac{n}{2} +b_{n-1}$
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{n-k+1}{2} + b_{n-k}$
Substitute k = n
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{1}{2} + b_{0}$
$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$
$b_n = \frac{n + n-1 +n-2 + \dots +1}{2}+0$
$b_n = \frac{n(n+1)}{4}$
$\frac{n(n+1)}{2} a_n = \frac{n(n+1)}{4}$
$a_n =\frac{1}{2}$
