I have the following sum to evaluate: $$\sum\limits_{n=1}^{\infty}\frac{n}{2^{n+1}}$$ My professor says that I can evaluate this through "differentiating the geometric series" but I am not quite sure what that means. I already know that this converges to $1$, just not sure how to prove.
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1See this. – David Mitra Dec 01 '17 at 17:37
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A simpler way: let $S=\sum_{n\geq 0}\frac{n}{2^{n+1}}$ (the series is obviously absolutely convergent). Then
$$ S = 2S-S = \sum_{n\geq 1}\frac{n}{2^n}-\sum_{n\geq 1}\frac{n}{2^{n+1}}=\sum_{n\geq 0}\frac{n+1}{2^{n+1}}-\sum_{n\geq 1}\frac{n}{2^{n+1}} $$ hence $$ S = \frac{1}{2}+\sum_{n\geq 1}\frac{1}{2^{n+1}} = \frac{1}{2}+\frac{1}{2}=\color{red}{1}.$$
Jack D'Aurizio
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Hint: recall that $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
differentiate both sides of this equation and multiply by $x$. Then substitute $x=\frac{1}{2}$.
In particular note that $$\frac{d}{dx} \sum_{n=0}^{\infty} x^n=\sum_{n=0}^\infty \frac{d}{dx} x^n=\sum_{n=0}^\infty n \cdot x^{n-1}$$ on the right hand side of the equation.
Andres Mejia
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