Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.
I tried to use the Ratio test:
$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$
$\lim\limits_{n \rightarrow \infty}{{(\frac{1}{2}(1+ \frac{1}{n+1})) = \frac{1}{2}}} \leq 1$
So $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges.
Could somebody please check my solution?

The series converges, it's can be simply proved finding it's upper bound or as you've done it by the ratio test. But if you are interested in finding it's actual value here's a way to do it. From the sum of geometric series we have:
$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$
Multiply both sides by $r$
$$\sum_{n=0}^{\infty} r^{n+1} = \frac{r}{1-r}$$
Differentiate with respect to $r$. Then we have:
$$\sum_{n=0}^{\infty} (n+1)r^n = \frac{1}{(r-1)^2}$$
Now substitute $r= \frac 12$. We have:
$$\sum_{n=0}^{\infty} \frac{n+1}{2^n} = \frac{1}{(-\frac 12)^2} = 4$$ 
