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This problem comes from another equation on another question (this one). I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided.

Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$

But that's not what I wanted. Could anyone help me?

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  • $\begingroup$ This is NOT a geometric series. $\endgroup$ – Anurag A Jun 14 '15 at 18:02
  • $\begingroup$ The mistake you've made is that $\sqrt k^k \neq k$ $\endgroup$ – ptrsinclair Jun 14 '15 at 18:03
  • $\begingroup$ $\sqrt{k}^k \neq k$ so that's where your last line of algebra went wrong $\endgroup$ – Zach Stone Jun 14 '15 at 18:03
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    $\begingroup$ This may help. $\endgroup$ – David Mitra Jun 14 '15 at 18:04
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    $\begingroup$ Possible duplicate of How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? $\endgroup$ – Hans Lundmark Apr 14 '18 at 15:07
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$$S=\sum_{k=0}^\infty{k\over2^k}=2\sum_{k=0}^\infty{k\over2^{k+1}}=2\sum_{k=1}^\infty{k-1\over2^k}=2S-2\sum_{k=1}^\infty{1\over2^k}=2S-2\\ \therefore S=2$$

To prove convergence of $S$ we can use ratio test...

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    $\begingroup$ This is only correct if you already know that $S$ exists (i.e. that the sum doesn't diverge). Otherwise, very nice! $\endgroup$ – Mathmo123 Jun 14 '15 at 18:15
  • $\begingroup$ Solid approach! +1 $\endgroup$ – Mark Viola Jun 14 '15 at 19:00
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Start with: $$\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k.$$ Then take derivative with respect to $x$. $$\frac{1}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k-1}.$$ Multiply by $x$. $$\frac{x}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k}.$$ Now substitute $x=\frac{1}{2}$.

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  • $\begingroup$ am I wrong? if you place $x=1/2$ then the summation will be negative. $\endgroup$ – kelalaka Nov 8 '18 at 20:33
  • $\begingroup$ @kelalaka I had a typo with a negative sign. It's fixed. Thanks for pointing it out. $\endgroup$ – Anurag A Nov 8 '18 at 21:34
  • $\begingroup$ I check that someone edited, and the community accepted. Your original answer was correct. Revering back is also possible. $\endgroup$ – kelalaka Nov 9 '18 at 8:16
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You can also view it this way, which is quite intuitive, though possibly not that rigor:

\begin{align*} \sum_{k=0}^\infty \frac{k}{2^k} &= \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\ &= \frac12 + \Bigl(\frac14 + \frac14\Bigr) + \Bigl(\frac18 + \frac18 + \frac18\Bigr) + \Bigl(\frac1{16} + \frac1{16} + \frac1{16}\Bigr) + \cdots \\ &= \Bigl(\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac18 + \frac1{16} + \cdots\Bigr) + \cdots \\ &= 1 + \frac12 + \frac14 + \frac18 + \cdots \\ &= 2. \end{align*}

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