compute the sum of a series $$
\sum_{n=0}^\infty \frac{x^n}{n(n+1)}
$$
what is the sum of this?
I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there

Your sum has an error in it because for n=0 you are dividing by 0. However, if you start instead with n=1, there is no error, so I will assume you meant that.
Notice that $$\int \int x^n dx dx = \frac{x^{n+2}}{(n+1)(n+2)} +Cx+D =x \cdot \frac{x^{n+1}}{(n+1)(n+2)} +Cx+D$$
Therefore $$\int \int \sum_{n=0}^{\infty} x^n dx dx = Cx+D+ \sum_{n=0}^{\infty} x \cdot \frac{x^{n+1}}{(n+1)(n+2)} =Cx+D+  x \cdot \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)(n+2)}$$
$$ =Cx+D+  x \cdot \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} $$. 
Letting $C = D = 0$ and dividing by $x$ gives a closed form expression for the sum. We also have that first integral, $$\int \int \sum_{n=0}^{\infty} x^n dx dx = \int \int \frac{1}{1-x} dx dx =\int C + \log(x-1) dx = Cx + D + (x-1) \log(x-1) $$
Again letting $C = D = 0$ and dividing by $x$, we have $$ \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \frac{(x-1)\log(x-1)}{x}  $$
