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There is this equation:

$$M = \sum\limits_{i = 1}^{\log n} {\frac{{in}}{{{2^i}}}} = n\sum\limits_{i = 1}^{\log n} {\frac{i}{{{2^i}}}} \leqslant n\sum\limits_{i = 1}^\infty {\frac{i}{{{2^i}}} = 2n} $$

And I don't understand why the rightmost summation can be simplified to 2n. Can you please explain it to me?

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  • 4
    $\begingroup$ We have the identity $\sum\limits_{k=1}^\infty kx^k=\frac{x}{(1-x)^2}$, which can be derived from differentiating the usual geometric series. Now, take $x=\frac12$... $\endgroup$ – J. M. is a poor mathematician Apr 24 '12 at 12:44
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Knowing that: $$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1 $$

We have: $$\begin{eqnarray*} \sum_{i=1}^\infty \frac{i}{2^i} &=& \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots \\ &=& (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{8} + \frac{1}{16} + \cdots) + \cdots \\ &=& 1 + \frac{1}{2} + \frac{1}{4} + \cdots \\ &=& 2 \end{eqnarray*}$$

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You need to know that:

sum ( i / 2^i ) = sum ( 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ... ) = 2

See: http://www.wolframalpha.com/input/?i=sum%28i%2F2%5Ei%29

(It's an arithmetic-geometric series.)

From there the rest should be obvious.

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