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I was searching for an answer to the sum of this serie but i can't understand what was done in the second step here

how can i go from this $\sum_{n=0}^\infty (n+1) 2^{-n}$

to this $ = \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$

this is part of the answer too: "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" $\sum_{n=0}^\infty n^3 3^{-n}$.

Mateo Fernandez
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    See here. – J.G. Nov 29 '22 at 14:33
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    Note $\sum\limits_{n=0}^\infty (n+1) 2^{-n} = \sum\limits_{n=1}^\infty n 2^{-n-1}$ and $\sum\limits_{n=0}^\infty (n+1) 2^{-n}=\sum\limits_{n=0}^\infty n 2^{-n} +\sum\limits_{n=0}^\infty 2^{-n}$ – Henry Nov 29 '22 at 14:34

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