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What is $ \sum_{n=1}^{\infty} \frac{n}{2^{n}}$ equal to?
I know by the ratio test this converges absolutely, thats not what I'm looking for in this case, I want the actual value, is there a way aside from numerical approximation to get this value?
The value obtained from the ratio test is $\frac{1}{2}$ now I just need to know what this approximates to, Wolfram says 2 but how would one prove this?

Thanks

Wallace
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  • use $\sum nx^n=x(\sum x^n)'$ for $x=\frac 12$. – zwim Nov 21 '20 at 20:54
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    Does this answer your question? Duplikate. https://math.stackexchange.com/q/337937/681678 – thinkingeye Nov 21 '20 at 20:54
  • you have the sum of $n x^n$ where you will eventually set $x = 1/2.$ Some care is needed. I would divide out a single $x$ (note the sum begins at $n=1$ ) to arrive at the sum of $n x^{n-1}$ – Will Jagy Nov 21 '20 at 20:56
  • Just as an aside, is there a probabilistic argument that $\sum^\infty_{n=0}\frac{n}{2^{n+1}}=1$? – Paul Nov 21 '20 at 21:07
  • @thinkingeye that did help thank you – Wallace Nov 21 '20 at 21:08
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    @Paul With a fair coin the expected number of tails before the first heads is $1$ – Hagen von Eitzen Nov 21 '20 at 21:12
  • Alternatively, an algebraic approach would be, $$\frac 12+\frac 24+\frac 38+\dots=(1/2+1/4+\dots)+(1/4+1/8+1/16+\dots)+(1/8+1/16+\dots)+\dots$$ Now, sum each of the geometric series to get $1+1/2+1/4+\dots=2$. The rearrangement of terms is valid since the series is absolutely convergent. – Prasun Biswas Nov 21 '20 at 21:16
  • @Hagen much nicer. – Paul Nov 21 '20 at 22:08

2 Answers2

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We can use a trick on the geometric series. $$\sum^\infty_{n=0}x^n=\frac{1}{1-x}$$ We can take the derivative with respect to $x$ of both sides $$\sum^\infty_{n=0}nx^{n-1}=\frac{1}{(1-x)^2}$$ Let $x=\frac{1}{2}$ $$\sum^\infty_{n=0}\frac{n}{2^{n-1}}=\frac{1}{(1-\frac{1}{2})^2}$$ $$\sum^\infty_{n=0}\frac{n}{2^n}=2$$

razivo
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Let's say we have $$ f(r) = \sum_{n=0}^{\infty}r^{n} $$ Of course, in order for this to make sense we require ${|r|<1}$. So our function is only defined on ${(-1,1)}$. Now, let's take the derivative: $$ \frac{df}{dr}=\frac{d}{dr}\sum_{n=0}^{\infty}r^n $$ Now here's the question: can you interchange the sum and derivative operator here? In other words, is it true that $$ \frac{d}{dr}\sum_{n=0}^{\infty}r^n =^{?} \sum_{n=0}^{\infty}\frac{d}{dr}r^n $$ in this case - the answer is yes. But this is not always true for a general series involving functions. I'd suggest researching the conditions necessary for doing such interchange for future reference, but for now you can take my word in this case it's okay to do so.

So we get $$ \frac{df}{dr}=\sum_{n=0}^{\infty}\frac{d}{dr}r^n = \sum_{n=0}^{\infty}nr^{n-1} $$ The cool thing is, we know an alternative representation for ${f(r)}$. That is $$ f(r) = \sum_{n=0}^{\infty}r^n = \frac{1}{1-r} $$ This tells us $$ \frac{df}{dr}=\sum_{n=0}^{\infty}nr^{n-1} = \frac{d}{dr}\left(\frac{1}{1-r}\right) $$ Using this knowledge, are you now able to show that $$ \sum_{n=0}^{\infty}n\frac{1}{2^n} = 2 $$ ?

Riemann'sPointyNose
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