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What is the right way to assess this problem?

$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$

To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. However it does not seem to be the case as there is no common ratio between the terms. Since the numerator is increasing from $1,2,3,...$ makes it impossible to find a ratio. What should I do?.

Chris Steinbeck Bell
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4 Answers4

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Hint: Try looking at $3E - E$.

Michael Biro
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  • I reached the answer by following your hint $E=\frac{3}{4}$ however I'm wondering how did you got to that observation?. – Chris Steinbeck Bell Oct 19 '17 at 15:00
  • @ChrisSteinbeckBell The numerators increase by $1$ every term, so if we could subtract from each numerator its predecessor, we'll get a series with all numerators equal to $1$, which is geometric. To subtract from the numerators, we just need to shift them up or down one term will give the exact cancellation we needed. This works out really nicely in a case like this because the shift is just changing the power of $3^n$ in the denominator, which is basically multiplying by $3$. So $3E - E$ is "Take the series with all the numerators shifted up, and subtract the original numerators." – Michael Biro Oct 19 '17 at 15:10
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You can split it up into a sum of geometric progressions: \begin{align} 3^{-1} + 2 \cdot 3^{-2} + 3 \cdot 3^{-3} + \dotsb = 3^{-1} + 3^{-2} + 3^{-3} + \dotsb \\ + 3^{-2} + 3^{-3} + \dotsb \\ + 3^{-3} + \dotsb \\ \ddots \\ = 1(3^{-1}+ 3^{-2} + 3^{-3} + \dotsb) \\ +3^{-1}(3^{-1}+ 3^{-2} + \dotsb) \\ + 3^{-2}(3^{-1} + \dotsb) \\ \ddots \\ = (1+3^{-1}+3^{-2}+\dotsb)(3^{-1}+3^{-2}+3^{-3}+\dotsb) \end{align}

Chappers
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  • The product of the two brackets does not seem to reproduce the original values. Can you check into that? – Chris Steinbeck Bell Oct 19 '17 at 15:07
  • @ChrisSteinbeckBell Ah yes, my index was off by 1. – Chappers Oct 19 '17 at 15:08
  • I got your idea, but Can you explain how did you got from the sum of geometric progressions to the two brackets?. This latter part is not very obvious to me. – Chris Steinbeck Bell Oct 19 '17 at 15:11
  • @ChrisSteinbeckBell I've added another line to the working. Clearer? – Chappers Oct 19 '17 at 15:31
  • Yes its clearer now, however I was not familiar with the procedure. Do you know some documentation or other sources from where I can check worked out examples on how this method is used?. – Chris Steinbeck Bell Oct 19 '17 at 15:40
  • In summation notation, it's rather easy to see what's happened: $$\sum_{k=1}^{\infty} k 3^{-k} = \sum_{k=1}^{\infty} \sum_{n=1}^k 3^{-k} = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} 3^{-k}, $$ so this is called changing the order of summation in its general form. Here there's a further simplification in that the internal sums can all be written as $3^{-n}\sum_{j=0}^{\infty} 3^{-j} $, and the remaining sum can be pulled out of the external sum as a constant factor. I would have thought just searching for "changing the order of summation" would produce plenty of examples. – Chappers Oct 19 '17 at 16:04
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\begin{eqnarray} E&=&\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots=\sum_{k=1}^\infty\frac{k}{3^k}=\frac{1}{3}\sum_{k=1}^\infty k\cdot\left(\frac{1}{3}\right)^{k-1}\\ &=&\frac{1}{3}\dfrac{d}{dx}\left(\sum_{k=0}^\infty x^k\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\frac{d}{dx}\left(\frac{1}{1-x}\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\dfrac{1}{(1-x)^2}\Big|_{x=\frac{1}{3}}\\ &=&\frac{1}{3}\cdot\dfrac{1}{\left(1-\frac{1}{3}\right)^2}=\frac{1}{3}\cdot\dfrac{1}{\frac{4}{9}}=\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4} \end{eqnarray}

HorizonsMaths
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1

$$\sum _{k=1}^{\infty } \frac{k}{3^k}=\frac34$$ To prove this I'll use generating functions

Consider $$f(x)=\sum _{k=1}^{\infty } \frac{ x^k}{3^k}=\sum _{k=1}^{\infty } \left(\frac{ x}{3}\right)^k=\frac{1}{1-x/3}=\frac{3}{3-x}\quad(*)$$

$$f'(x)=\frac{1}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}$$

To make the index back to $k$ multiply both sides by $x$ so that

$$xf'(x)=\frac{x}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}=\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k}$$ The sum is then $xf'(x)$ for $x=1$

$$\sum _{k=1}^{\infty } k\,\frac{1}{3^k}$$

On the other side $(*)$ we have $f'(x)=\dfrac{3}{(3-x)^2}$

that is $xf'(x)=\dfrac{3 x}{(3-x)^2}$

The sum $\sum _{k=1}^{\infty } \frac{k}{3^k}$ is $xf'(x)$ for $x=1$, which gives $\dfrac{3}{4}$

Raffaele
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  • Why the sum is $xf'(x)$?. I see your approach uses calculus and it took me a while to recall this, however I'm stuck at trying to understand the second line since $\frac{x}{3}$ cancels out with $(\frac{x}{3})^{k-1}$ therefore returning to the value of the first line. But that wouldn't mean $k\frac{3}{3-x}$? How does the later becomes into $\frac{3x}{(3-x)^{2}}$?. I'm sorry but this is not very clear to me. – Chris Steinbeck Bell Oct 19 '17 at 15:52
  • @ChrisSteinbeckBell have just edited and added more explanation – Raffaele Oct 19 '17 at 16:37
  • Okay I understood the additional steps you added but I'm still stuck between the little star and $\frac{3x}{(3-x)^{2}}$ It tells "other side", but not sure which side on $f(x)$ to look for. – Chris Steinbeck Bell Oct 20 '17 at 03:50