3
$\begingroup$

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?

$\endgroup$
  • 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. isn't a mathematician Apr 24 '12 at 12:44
7
$\begingroup$

Knowing that: $$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1 $$

And hence by removing successive terms from the left:

$$ \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{2} $$ $$ \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{4} $$ $$ \vdots $$

We have: $$\begin{eqnarray*} \sum_{i=1}^\infty \frac{i}{2^i} &=& \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots \\ &=& (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{8} + \frac{1}{16} + \cdots) + \cdots \\ && \text{Substituting the identites from above:} \\ &=& 1 + \frac{1}{2} + \frac{1}{4} + \cdots \\ && \text{And substituting the first identity again:} \\ &=& 1 + 1 \\ &=& 2 \end{eqnarray*}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.