<s>[INST] Convergence of $\sum\limits_{n=1}^\infty \frac {n+1}{2^n}$. Test Convergence of $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n}$$
Attempt: $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n} = \sum\limits_{n=1}^\infty \dfrac {n }{2^n} + \sum\limits_{n=1}^\infty \dfrac {1}{2^n}$$
The second summation is definitely convergent. So, we need to just investigate if the first summation is convergent or not. 
Let $$X = \sum\limits_{n=1}^\infty \dfrac {n }{2^n}$$
Is there a way to test convergence of this summation without the integral test?
Thank you very much for your help.
 [/INST] Let:
$$ S_N = \sum_{n=1}^{N}\frac{n+1}{2^n}. $$
We have:
$$\begin{eqnarray*} \frac{S_N}{2}&=&S_N-\frac{S_N}{2}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=1}^{N}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=2}^{N+1}\frac{n}{2^{n}}\\&=&1+\sum_{n=2}^{N}\frac{1}{2^n}-\frac{N+1}{2^{N+1}}=\frac{3}{2}-\frac{N+3}{2^{N+1}}.\end{eqnarray*}$$
Since $\frac{N+2}{2^{N+1}}\to 0$ as $N\to +\infty$, we have $S_N\to\color{red}{3}$.
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