How to calculate this $\sum\limits_{n=0}^{\infty}\frac{n}{2^n}e^{-jwn}$,because it is not geometric progression,so i can't know how to solve it,can anyone help me?
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Still you can write your series as $$\sum_{n\geq 0}\frac{e^{-jwn}}{2^n}+\sum_{n\geq 1}\frac{e^{-jwn}}{2^n}+\sum_{n\geq 2}\frac{e^{-jwn}}{2^n}+\ldots $$ which is a geometric series of geometric series. – Jack D'Aurizio Dec 04 '18 at 10:43
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See https://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn – lab bhattacharjee Dec 04 '18 at 10:43
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For $|z|<1$ we have $\sum_{n=0}^{\infty}n z^n= \frac{z}{(1-z)^2}$.
Now let $z= \frac{e^{-jw}}{2}$. Can you proceed ?
Fred
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thx,i will see this https://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn – electronic component Dec 04 '18 at 10:47