# Sum of $\sum\limits_{n=0}^{\infty} (2n+1) (\frac{1}{2})^n = 1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+…$ [duplicate]

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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## 2 Answers

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?

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.

• I think you've meant $kx^{k-1}$ :) – Evgeny Jan 22 '16 at 12:59
• I only gave an input idea, it's up to the OP to sort out the algebra. – Alex Jan 22 '16 at 13:01
• 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. – Evgeny Jan 22 '16 at 13:31
• $2\sum_k k x^k + \sum_k x^k$ so there's no mistake – Alex Jan 23 '16 at 17:32