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I've been dealing with the following series for a while now, without real progress.

$$\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$$

After using WolframAlpha, I know it converges to $30$, but I can't see how to calculate it by myself.

Any leads would be greatly appreciated!


marked as duplicate by lab bhattacharjee sequences-and-series Mar 2 '18 at 9:41

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  • $\begingroup$ There's one more naive method other than good answer already given, i was talking about denoting it as S and subtracting from it 2 thirds of S. Again repeating similar step. It requires sum to be convergent $\endgroup$ – King Tut Mar 2 '18 at 8:43
  • $\begingroup$ @KingTut Do you mean to repeat the subtraction infinitely? $\endgroup$ – GalAbra Mar 2 '18 at 9:00
  • $\begingroup$ No, I said twice. $\endgroup$ – King Tut Mar 2 '18 at 10:12
  • $\begingroup$ See math.stackexchange.com/a/594008/520782 $\endgroup$ – King Tut Mar 2 '18 at 10:14

Consider the function $f$ defined by $f(x)=\sum_{n=0}^\infty x^n=\frac1{1-x}$ (if $|x|<1$). Then$$f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2}=\frac1{x^2}\left(\sum_{n=2}^\infty n^2x^n-\sum_{n=2}^\infty nx^n\right).$$Can you take it from here?

  • $\begingroup$ Should (n^2)(x^2) be (n^2)(x^n)? Or did I misunderstand something? $\endgroup$ – GalAbra Mar 2 '18 at 8:57
  • $\begingroup$ @GalAbra I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Mar 2 '18 at 8:57

For $-1<x<1$, $\displaystyle \sum_{n=1}^{k}x^n=\frac{x(1-x^{k})}{1-x}$.


\begin{align*} \sum_{n=1}^knx^{n-1}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\\ &=\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}\\ \end{align*}


\begin{align*} \sum_{n=1}^knx^{n}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\\ &=\frac{x-(k+1)x^{k+1}+kx^{k+2}}{(1-x)^2}\\ \end{align*}

Then differentiate again and multiply by $x$. Put $x=\frac{2}{3}$. Take $k\to\infty$.


Substitute $2/3$ by a variable $x$. Then the series defined by

$$\sum_{k=1}^\infty k^2 x^k$$

converges locally uniformly for $|x|<1$, hence we can integrate and differentiate it termwise so

$$\sum_{k=1}^\infty k^2 x^k=x\sum_{k=1}^\infty k^2 x^{k-1}=x\sum_{k=1}^\infty k\frac{d}{dx}( x^k)=x\frac{d}{dx}\left(\sum_{k=1}^\infty kx^k\right)\\=x\frac{d}{dx}\left(x\sum_{k=1}^\infty k x^{k-1}\right)=x\frac{d}{dx}\left(x\sum_{k=1}^\infty \frac{d}{dx}(x^k)\right)=x\frac{d}{dx}\left(x\frac{d}{dx}\sum_{k=1}^\infty x^k\right)=\left[x\frac{d}{dx}\right]^2\frac{x}{1-x}\\=\left[x\frac{d}{dx}\right]\frac{x}{(1-x)^2}=x\left(\frac{1}{(1-x)^2}+\frac{2x}{(1-x)^3}\right)=\frac{x(1+x)}{(1-x)^3}$$


$$\sum_{k=1}^\infty k^2(2/3)^k=\left[\sum_{k=1}^\infty k^2 x^k\right]_{x=2/3}=\frac{x(1+x)}{(1-x)^3}\bigg|_{x=2/3}=\frac{10}9\cdot 3^3=3\cdot 10=30$$

  • Observe that $f(x)=\sum_{n=0}^{\infty}x^n=(1-x)^{-1}$ if $|x|<1$
  • On both sides take the derivative: $\sum_{n=1}^{\infty}nx^{n-1}=(1-x)^{-2}$
  • Again on both sides take the derivative: $\sum_{n=2}^{\infty}n(n-1)x^{n-2}=2(1-x)^{-3}$

Based on the LHS summations you can find an expression for $\sum_{n=1}^{\infty}n^2x^n$ and substitute $x=\frac23$.

I leave that to you.


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