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I'm having trouble to show that: $$ \sum_{n=1}^{\infty} \frac {2n(-1)^{n}}{4^{n+1}} = -2/25 $$

What I've tried so far: simplified the expression to: $$ \sum_{n=1}^{\infty} \frac{n}{2} (\frac{-1}{2})^{n} $$ I've tried to look for anything that looks like a power series to solve the problem, but the n in the numerator troubles me to find that.

Any help is appreciated!

moonshine2
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  • $ \sum_{n=1}^{\infty} \frac {2nx^{n}}{4^{n+1}} ; ;$ converges for $|x| < 4$ – Will Jagy Sep 20 '22 at 22:57
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Sep 20 '22 at 23:06
  • Yes, but in this question when I'm trying to apply the limit the n in the numerator troubles me to find for what value it converges to. Any tips in how to show that? – moonshine2 Sep 20 '22 at 23:06
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    @Shaun Oh! Thanks! I'll edit a little the question! – moonshine2 Sep 20 '22 at 23:06
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    @Shaun hope it's better! – moonshine2 Sep 20 '22 at 23:19
  • Yes, that's better. – Shaun Sep 20 '22 at 23:24
  • You need the formula for $n\cdot x^n$ . The trick is to derive the geometric sum with $x^n$ which almost gives the desired expression. Try to work it out. – Peter Sep 21 '22 at 06:15
  • @Peter Thanks for the answer! After someone linked the related "already has answers here" I did the question in 3 different ways, and one is by doing the derivation! Thanks for the reply. – moonshine2 Sep 21 '22 at 13:54

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