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I am learning about series of numbers at the moment. In the book there is an exercise in which I need to find the sum of : $$\sum_{i=1}^{\infty}\frac{i}{2^i}$$ I know it is equal to $2$. But how do I get to that result? Are there any general ways of finding the sums of series? In the books I am using, there is a lot about series, their convergence etc. but almost no examples.

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Let

$$S := \sum_{k=1}^\infty \frac{k}{2^k}.$$

Since all of its elements are positive and the series is absolutely convergent, we can do the following:

\begin{align*} S &= \sum_{k=1}^\infty \frac{k}{2^k} = \sum_{k=1}^\infty \frac{k-1}{2^k} + \sum_{k=1}^\infty \frac{1}{2^k} = \frac{1}{2} \sum_{k=1}^\infty \frac{k-1}{2^{k-1}} + 1 = \frac{1}{2} \sum_{k=2}^\infty \frac{k-1}{2^{k-1}} + 1 = \frac{1}{2} \sum_{k=1}^\infty \frac{k}{2^k} + 1 \\ &= \frac{1}{2}S + 1, \end{align*}

so $S = 2$.

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In this particular case:

$$\sum_{i=1}^{\infty}\frac{i}{2^i}=\sum_{i=1}^{\infty}\frac{1}{2^i}+\sum_{i=2}^{\infty}\frac{1}{2^i}+\sum_{i=3}^{\infty}\frac{1}{2^i}+\dots=\sum_{i=1}^{\infty}\frac{1}{2^i}\sum_{j=0}^{\infty}\frac{1}{2^j}=1\cdot 2$$.

But, of course, you need to say something about absolute convergence to manipulate the series in such a way.

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  • $\begingroup$ Perhaps you should use two different indices in your last expression. $\endgroup$ – Cameron Williams Feb 20 '14 at 19:16
  • $\begingroup$ They are "independent" but I agree, that looks better and less confusing. Fixed. $\endgroup$ – Vadim Feb 20 '14 at 19:18
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Let $x=\frac{1}{2}$. Then we have $\sum_{n=1}^{\infty}nx^n=x\sum_{n=1}^{\infty}nx^{n-1}$. Can you see a derivative under a sum?

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