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Could someone help me sum following series as $n$ goes to infinity: $$\sum^n_{i=1} \frac i {2^i} = \frac 1{2^1} + \frac 2{2^2} + \frac 3{2^3} +\cdots+ \frac n{2^n}$$

I don't need the answer so much as being pointed towards what I need to know to figure out the answer...I assume there are different ways to do it. I used to know this stuff a long time ago but don't know where to begin now. Thanks...

jimjim
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Ben S.
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2 Answers2

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Hint: Use the fact that $\dfrac{n}{x^n} = -x \cdot \dfrac{d}{dx}\left(\dfrac{1}{x^n}\right)$ to evaluate $\displaystyle \sum_{n=1}^{\infty} \frac{n}{x^n}$, and then substitute $x=2$.

Clive Newstead
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    Well, I can see that n/x^n=−x⋅(d/dx)[(1/x^n)] is true, but I"m afraid I don't know how to proceed when I sub that into the summation expression. If I distribute the summation across -x and (d/dx)(1/x^n) does the summation work like an integral and cancel out the (d/dx)? My calculus is pretty shaky right now! – Ben S. Oct 05 '14 at 22:43
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    No, but you can interchange the summation symbol and the differential operator to get $$\sum_{n=1}^{\infty} \frac{n}{x^n} = \sum_{n=1}^{\infty} \left( -x \cdot \dfrac{d}{dx} \left( \frac{1}{x^n} \right) \right) = -x \cdot \dfrac{d}{dx} \left( \sum_{n=1}^{\infty} \frac{1}{x^n} \right)$$ – Clive Newstead Oct 05 '14 at 22:46
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    Ah. Thanks....think I got it. In general, how do you make the first step? I mean, how do you "see" that n/x^n can be replaced by the expression -x * d/dx(1/x^n)? I can tell it's true once you suggested it, but how does one know to start there? – Ben S. Oct 05 '14 at 22:49
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    Lots of practice. – Jeff Oct 05 '14 at 22:54
  • My thought was: there's an awkward $n$ multiplying everything, we might be able to get rid of it by looking at the summand as a derivative. (And the reason I thought this was... lots of practice.) – Clive Newstead Oct 05 '14 at 23:55
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Easier hint: $$ \begin{array}{ccll} \sum_1^\infty\frac{n}{x^n} &=& \frac{1}{x^1} + & \frac{1}{x^2} + &\frac{1}{x^3} + \ldots \\ &+& & \frac{1}{x^2} +& \frac{1}{x^3} + \ldots \\ &+& & & \frac{1}{x^3} + \ldots \\ &+& \cdots \end{array} $$ Each of these is a simple geometric series (for example, row 2 sums to $\frac{1}{x^2} \frac{1}{1-1/x}$).

But each of those series is $1/x$ times the one above it. So you have a second geometric sum to do.

Mark Fischler
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