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I know this is a series of an arithmetic and geometric progression product which looks like

$$\sum\limits_{n=0}^{\infty} (2n+1) \left(\frac{1}{2}\right)^n$$

but I don't know how to calculate the sum. Any help would be appreciated!

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marked as duplicate by lab bhattacharjee calculus Jan 22 '16 at 13:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Consider $$f(x)=\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ $$x f'(x)=\sum_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}$$ Put these together and put in $x=\frac12$. Can you proceed on your own?

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Split into two sums, of the form $\sum_k k x^k + \sum_k x^k$. The first can be solved by identifying a derivative and interchanging it with the sum, the second is a convergent Geometric series.

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  • $\begingroup$ I think you've meant $kx^{k-1}$ :) $\endgroup$ – Evgeny Jan 22 '16 at 12:59
  • $\begingroup$ I only gave an input idea, it's up to the OP to sort out the algebra. $\endgroup$ – Alex Jan 22 '16 at 13:01
  • $\begingroup$ Well, okay, your answer supposed to be a hint rather than a full solution, but leaving an obvious mistake seems controversial to me even in the hint. $\endgroup$ – Evgeny Jan 22 '16 at 13:31
  • $\begingroup$ $2\sum_k k x^k + \sum_k x^k$ so there's no mistake $\endgroup$ – Alex Jan 23 '16 at 17:32

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