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$$ \frac{1}{x}+\frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \cdots =\frac{x}{(x-1)^2} $$

I can feel it. I can't prove it. I have tested it, and it seems to work. Domain-wise, I think it might be $x>1$, the question doesn't specify. Putting the LHS into Wolfram Alpha doesn't generate the RHS (it times out).

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What did you put into Wolfram Alpha? Please provide a link :) –  Shaun Jun 10 at 13:56
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m.wolframalpha.com/input/… I was looking to see if it would give the infinite sum as in the case of m.wolframalpha.com/input/… –  Orange Peel Jun 10 at 14:01
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This is a polylogarithm. –  Lucian Jun 10 at 15:02

5 Answers 5

up vote 12 down vote accepted

Consider $$ \sum_{n=0}^\infty y^n=\frac{1}{1-y}\quad;\quad\text{for}\ |y|<1.\tag1 $$ Differentiating $(1)$ with respect to $y$ yields $$ \sum_{n=1}^\infty ny^{n-1}=\frac{1}{(1-y)^2}.\tag2 $$ Multiplying $(2)$ by $y$ yields $$ \sum_{n=1}^\infty ny^{n}=\frac{y}{(1-y)^2}.\tag3 $$ Now plug in $y=\dfrac1x$ where $|x|>1$ to $(3)$ yields $$ \large\color{blue}{\sum_{n=1}^\infty \frac{n}{x^n}=\frac{x}{(x-1)^2}}. $$

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Note that your last step requires $x>0$. –  Martin Argerami Jun 10 at 13:54
    
But $y<1$, so $x>1$. –  Mark Hurd Jun 10 at 15:22
    
@MarkHurd Indeed, you're correct. I edited it. –  Tunk-Fey Jun 10 at 15:25

For $\;|x|<1\;$ :

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies \frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies$$

$$\implies\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n$$

So no: you don't have an equality there...

Added on request: If $\;|x|>1\;$ then we can do:

$$\frac x{(x-1)^2}=\frac1x\frac1{\left(1-\frac1x\right)^2}=\frac1x\left(\sum_{n=0}^\infty\frac1{x^n}\right)^2$$

So still not the same expression as in the question.

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Dang you got quick fingers :D –  DanZimm Jun 10 at 13:37
    
Wow. Dead end there. Thanks anyway! –  Orange Peel Jun 10 at 13:38
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But what if $|x|>1$? It does seem true in that case, which is what the OP asked in the question. –  Nicholas Stull Jun 10 at 13:43
    
I added some stuff that will possibly help you, @OrangePeel –  DonAntonio Jun 10 at 13:53
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@DonAntonio: if you replace $x$ with $1/x$ in your last expression, you get the OP's formula. –  Martin Argerami Jun 10 at 13:56

Let's put this together cleanly:

For the sum to converge, $|x| > 1$ which means $$\sum_{n=0}^\infty x^{-n} = \frac{x}{x-1}$$

Squaring this gives $$\frac{x^2}{(x-1)^2} = \sum_{n=0}^\infty \sum_{m=0}^\infty x^{-(n+m)} $$ $$ = \sum_{r=0}^\infty (r+1)x^{-r}$$

(because each value of $(n+m)$ occurs $(n+m+1)$ times in the infinite sum (0 = 0+0, 1 = 1+0 or 0+1, 2=0+2 or 1+1 or 2+0, and so on).

Divide by $x$ gives:

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

$$ = \frac1{x} + \frac{2}{x^2} + \frac{3}{x^3} + ... $$

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HINT: for $|y|<1$

$$\sum_{n=0}^{\infty} n y^n=y\cdot\frac{d \sum_{n=0}^{\infty} y^n}{dy}=y\cdot\frac{d\left(\dfrac1{1-y}\right)}{dy}=\frac y{(1-y)^2}$$

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I think a less formal solution could be more understandable.

consider $$ S_n= \frac{1}{x} + \frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \dots + \frac{n}{x^n}$$

$$ xS_n = 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \dots + \frac{n}{x^{n-1}}$$ then $$xS_n - S_n = 1+ (\frac{2}{x}-\frac{1}{x})+(\frac{3}{x^2}-\frac{2}{x^2})+\dots+(\frac{n}{x^{n-1}}-\frac{n-1}{x^{n-1}}) - \frac{n}{x^n}$$ $$S_n(x-1) = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}} - \frac{n}{x^n}$$ Now we have a simplified the problem to one of a basic geometric series, so $$S_n(x-1) = T_{n-1} - \frac{n}{x^n}$$ where $$T_{n-1} = 1 + \frac{1}{x} + \frac{1}{x^2}+\dots+\frac{1}{x^{n-1}}$$ $$\frac{T_{n-1}}{x} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}+\dots+\frac{1}{x^{n}}$$ $$T_{n-1} - \frac{T_{n-1}}{x} = 1 + (\frac{1}{x}-\frac{1}{x})+ (\frac{1}{x^2}-\frac{1}{x^2})+\dots - \frac{1}{x^n}$$ $$T_{n-1}(1-\frac{1}{x}) = 1 - \frac{1}{x^n}$$ $$T_{n-1}(\frac{x-1}{x}) = \frac{x^n-1}{x^n}$$ $$T_{n-1} = \frac{x^n-1}{x^n}\cdot(\frac{x}{x-1})$$ $$T_{n-1} = \frac{x^n-1}{x-1}\cdot(\frac{1}{x^{n-1}})$$ $$T_{n-1} = \frac{x-\frac{1}{x^{n-1}}}{x-1}$$ Thus $S_n(x-1)$ becomes $$S_n(x-1) = \frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$ for $|x|\gt 0$ this becomes $$\lim_{n\to\infty}S_n(x-1) = \lim_{n\to\infty}\frac{x-\frac{1}{x^{n-1}}}{x-1} - \frac{n}{x^n}$$ $$S(x-1) = \frac{x-\displaystyle\lim_{n\to\infty}\frac{1}{x^{n-1}}}{x-1} - \lim_{n\to\infty}\frac{n}{x^n}$$ $$S(x-1) = \frac{x-0}{x-1} - 0 = \frac{x}{x-1}$$ $$S = \frac{x}{(x-1)^2}$$ I used l'Hopital's rule to evaluate $\displaystyle\lim_{n\to\infty}\frac{n}{x^n}$, being an $\frac{\infty}{\infty}$ indeterminate form

This helps me to understand the problem. Afterwards, I would go on to compose a more formal proof.

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