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$$\sum_{n=1}^{\infty}n^210^{-n} = \frac{110}{3^6}$$ I noticed this while playing around on my calculator. Is it true and how come?

Marijn
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    Are you familiar with power series? Namely, $$\sum_{n=1}^\infty r^{n-1}=\dfrac{1}{1-r}$$if $|r|<1$ (in this case, $r=\frac{1}{10}$, and you can deal with the $n^2$ term by considering the derivatives of the power series). – user170231 Dec 09 '15 at 21:34
  • @user170231 No not familiar with power series, but I'll look into it.. Thanks! – Marijn Dec 09 '15 at 21:37

2 Answers2

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$$\sum_{n=1}^{\infty} n^2 x^n = x \frac{d}{dx} \left [x \frac{d}{dx} \left (\frac1{1-x} \right ) \right ] $$

because

$$\frac1{1-x} = \sum_{n=0}^{\infty} x^n$$

Evaluating the derivatives, we get

$$\sum_{n=1}^{\infty} n^2 x^n = \frac{x(1+ x)}{(1-x)^3}$$

Plug in $x=1/10$ and the expected answer results.

Ron Gordon
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  • Does this method have a name. – Ahmed S. Attaalla Dec 09 '15 at 22:22
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    @AhmedS.Attaalla: not that I am aware of. However, experience tells us that increasing powers of $n$ in the sum are results of taking derivatives of the function represented by the sum, so long as the resulting sum converges. – Ron Gordon Dec 09 '15 at 22:24
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This is because: \begin{align*} \sum_n n^2x^n&=x^2\sum_n n(n-1)x^{n-2}+x\sum_n nx^{n-1}\\ &=x^2\Bigl(\frac 1{1-x}\Bigr)''+x\Bigl(\frac 1{1-x}\Bigr)'\\ &=\frac{2x^2}{(1-x)^3}+\frac x{(1-x)^2}=\frac{x^2+x}{(1-x)^3}=\color{red}{\frac{x(x+1)}{(1-x)^3}}. \end{align*} Then set $x=\dfrac1{10}$.

Bernard
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