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Calculate $\sum\limits_{x=0}^{\infty} \dfrac{x}{2^x}$

So, this series converges by ratio test. How do I find the sum? Any hints?

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marked as duplicate by Martin Sleziak, Claude Leibovici, LutzL, Henrik, Daniel Fischer Nov 19 '16 at 12:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Let the sum be $S$. Then $2S-S=\sum_{x=0}^\infty \frac{x+1-x}{2^x}$. $\endgroup$ – André Nicolas Jan 13 '15 at 3:32
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    $\begingroup$ We have $S=\frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots$. So $2S=\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\cdots$. Subtract, "lining up" powers of $2$. We get $2S-S=\frac{1-0}{1}+\frac{2-1}{2}+\frac{3-2}{4}+\frac{4-3}{8}+\cdots$. This sum is $2$ (geometric progression) so $S=2$. $\endgroup$ – André Nicolas Jan 13 '15 at 3:45
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    $\begingroup$ My version leaves out the first term (obviously harmless) and cancels a $2$ from top and bottom for all other terms. $\endgroup$ – André Nicolas Jan 13 '15 at 3:56
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    $\begingroup$ Quite so: alas, mathematical expressions aren't "searchable", so locating duplications mostly relies on potential respondents' memories of where they saw or replied to an earlier appearance of the question. (I know I've said something about this sum before -- some time ago...) $\endgroup$ – colormegone Jan 13 '15 at 19:55
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As a first step, let us prove that

$$f(r) := \sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

if $r \in (-1,1)$. This is the geometric series. If you haven't seen this proven before, here's a proof. Define

$$S_N = \sum_{n=0}^N r^n.$$

Then

$$r S_N = \sum_{n=0}^N r^{n+1} = \sum_{n=1}^{N+1} r^n = S_N - 1 + r^{N+1}.$$

Solve this equation for $S_N$, obtaining

$$S_N = \frac{1-r^{N+1}}{1-r}$$

and send $N \to \infty$ to conclude.

The sum above converges absolutely, so we can differentiate term by term. Doing so we get

$$f'(r) = \sum_{n=0}^\infty n r^{n-1} = \frac{1}{(1-r)^2}.$$

(Precisely speaking, the sum in the middle is ill-defined at $r=0$, in that it has the form $0/0$. However, $f'(0)=1$ still holds. This doesn't matter for this problem, but it should be noted regardless.) Now multiply by $r$ to change it into your form:

$$\sum_{n=0}^\infty n r^n = \frac{r}{(1-r)^2}.$$

Now substitute $r=1/2$.

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  • $\begingroup$ why does r = $\frac{1}{2}$ ? $\endgroup$ – 123 Jan 13 '15 at 3:40
  • $\begingroup$ Wow .. Thanks for such an awesome answer $\endgroup$ – square_one Jan 13 '15 at 3:43
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    $\begingroup$ @mathtastic With $r=1/2$ the summand is $n (1/2)^n = n/2^n$, as desired. $\endgroup$ – Ian Jan 13 '15 at 3:43
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If you write out the first few (non-zero) terms, $ \ \frac{1}{2^1} \ + \ \frac{2}{2^2} \ + \ \frac{3}{2^3} \ + \ \ldots \ $ , another approach suggests itself, which is to use a "stacking" of infinite series (this would be along the lines of Jacob Bernoulli-style [1680s] ) :

$$ \frac{1}{2^1} \ + \ \frac{2}{2^2} \ + \ \frac{3}{2^3} \ + \ \ldots \ \ = $$

$$ \frac{1}{2^1} \ + \ \frac{1}{2^2} \ + \ \frac{1}{2^3} \ + \ \ldots $$ $$ \quad \quad \quad \quad \ \ \ + \ \frac{1}{2^2} \ + \ \frac{1}{2^3} \ + \ \frac{1}{2^4} \ + \ \ldots $$ $$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ + \ \frac{1}{2^3} \ + \ \frac{1}{2^4} \ + \ \frac{1}{2^5} \ + \ \ldots \ \ = $$

$$ 1 \ + \ \frac{1}{2} \ + \ \frac{1}{4} \ + \ \frac{1}{8} \ + \ \ldots \ = \ \ 2 \ \ . $$

[summing the sub-series in each row individually]

More modern methods (such as described by Ian) are more elegant, but Bernoulli got a lot of "mileage" out of this approach and extensions of it back then.

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  • $\begingroup$ In modern notation this looks like: $\sum_{n=0}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty \sum_{m=1}^n \frac{1}{2^n} = \sum_{m=1}^\infty \sum_{n=m}^\infty \frac{1}{2^n}$. The clever trick is the interchange of order, which is again an application of the absolute convergence. From a problem-solving perspective, the hard part is indexing the "triangular array" correctly! $\endgroup$ – Ian Jan 13 '15 at 19:04

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