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
 A: $$
\sum\limits_{n=0}^{\infty}x^n=\frac1{1-x}
$$
Integrate once to obtain
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log\left({1-x}\right)$$
Integrate again to obtain 
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+1)(n+2)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}=\frac{\left( 1-x \right)\log  \left( 1-x \right)}x  +1$$
A: You start with $$\sum_{n=0}^{\infty} x^n=\frac 1{1-x}$$  Now if you integrate it you get $$\int \sum_{n=0}^{\infty} x^n\;dx=\sum_{n=0}^{\infty} x^{n+1} / (n+1)=\int \frac 1{1-x}\; dx$$  You have to deal with the $n$ in the denominator-this should be a clue.
A: Let $f(x)=\sum_{n=0}^\infty\frac{x^n}{n(n+1)}$ for $|x|<1$. Inside its radius of convergence we may differentiate a power series term by term, so
$$f'(x)=\sum_{n=1}^\infty\frac{x^{n-1}}{n+1}=\frac1{x^2}\sum_{n=2}^\infty\frac{x^n}{n}=\frac{\log(1-x)-1}{x^2}.$$
Integrating, we have
$$f(x)=f(0)+\int_0^x\frac{\log(1-t)-1}{t^2}\,\mathrm dt$$
and so you can calculate $f$ using integration by parts.
A: 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 $$ \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \frac{(x-1)\log(x-1)}{x}  $$
A: (Sum should be from $n=1$.)
$$\frac1{n(n+1)} = \frac1{n}-\frac1{n+1} $$
so that 
$$\begin{align}\sum_{n=1}^{\infty} \frac{x^n}{n (n+1)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1} \\ &= -\log{(1-x)} - \frac1{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}\\ &=-\log{(1-x)} - \frac1{x} \left [-\log{(1-x)}-x \right ]\\ &= 1+\left ( \frac1{x}-1\right )\log{(1-x)} \end{align}$$
Note that the radius of convergence is $1$, i.e., the sum only converges when $|x| \le 1$.
