<s>[INST] 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?
 [/INST] $\sum_{n=0}^\infty (n+1) 2^{-n}$
$ = \sum_{n=0}^\infty \left( n 2^{-(n-1)} - (n+1) 2^{-n} + 2 \cdot 2^{-n} \right)$
$ = \left( 0 \cdot 2^{-(0-1)} - \lim_{n \to \infty} (n+1)2^{-n} \right) + 2 \sum_{n=0}^\infty { 2^{-n} }$
$ = 4$
The first step can be found be guesswork in simple cases, or be systematically doing "anti-difference by parts" in almost the same manner as "integration by parts", where the "by-parts" formula can be derived in exactly the same way from the "product rule" for the difference operator. Either of the forward or backwards difference operators work. This can easily handle things like $\sum_{n=0}^\infty n^3 3^{-n}$.
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