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I'm wondering how to prove the following series:

$$\alpha^2 \sum_{n=1}^\infty n^2 (1-\alpha)^{(n-1)} = \frac{2-\alpha}{\alpha}$$

user
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Mohammad Namakshenas
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  • Maybe the downvote was given because you haven't shown your work. If you can, try to elaborate your own solution and add here the details. We'll check that to give some more advice if needed. Good work! Bye – user Oct 31 '19 at 08:54
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    https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6 – lab bhattacharjee Oct 31 '19 at 08:54
  • You can also find many worked examples here on MSE. Try to find that also using Approach0. – user Oct 31 '19 at 08:55

1 Answers1

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The idea is to use that for $|x|<1$

$$\frac{d}{dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$$

$$\frac{d}{dx} \sum_{n=1}^\infty nx^{n-1}=\sum_{n=1}^\infty n(n-1)x^{n-2}$$

and some manipulation for geometric series.

user
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