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This question already has an answer here:

How to evaluate the sum of the series: $$\sum_{n=0}^{\infty}(n \cdot 3^n\cdot x^n)$$

I have tried with integration and derivation, but that doesn't lead to any intuitive (known) series.

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marked as duplicate by dxiv, Leucippus, Shailesh, Daniel W. Farlow, R_D Jan 7 '17 at 3:11

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$$S:=\sum_{n=0}^\infty n(3x)^n=\sum_{n=1}^\infty n(3x)^n=\sum_{n=0}^\infty (n+1)(3x)^{n+1}=3x\left(S+\sum_{n=0}^\infty (3x)^n\right) =3x\left(S+\frac1{1-3x}\right).$$

You can deduce $$S=\frac{3x}{(1-3x)^2}.$$

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Denote $H=\displaystyle\sum_{n=0}^{+\infty}nr^n$ whih $|r|<1.$ Then: $$H=r+2r^2+3r^3+4r^4+\cdots,$$ $$rH=r^2+2r^3+3r^4+4r^5+\cdots,$$ $$H-Hr=H(1-r)=r+r^2+r^3+\cdots=-1+(1+r+r^2+r^3+\cdots)$$ $$=-1+\frac{1}{1-r}=\frac{r}{1-r}\Rightarrow H=\frac{r}{(1-r)^2}.$$ For $r=3x,$ $$\sum_{n=0}^{+\infty}n3^nx^n=\frac{3x}{(1-3x)^2},\quad (|x|<1/3).$$

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Here is a variation using differentiation (see comment from @BrianMScott)

We obtain for $|x|<\frac{1}{3}$ \begin{align*} \sum_{n=0}^\infty n3^nx^n&=\sum_{n=1}^\infty n\left(3x\right)^n\tag{1}\\ &=3x\sum_{n=1}^\infty n(3x)^{n-1}\tag{2}\\ &=x\frac{d}{dx}\left(\sum_{n=1}^\infty(3x)^n\right)\tag{3}\\ &=x\frac{d}{dx}\left(\frac{1}{1-3x}-1\right)\tag{4}\\ &=\frac{3x}{(1-3x)^2} \end{align*}

Comment:

  • In (1) we start with $n=1$ since the term with $n=0$ is zero.

  • In (2) we factor out $3x$.

  • In (3) we use the differential operator and respect the chain rule (factor $3$).

  • In (4) we use the geometric series formula and subtract $1$ since the index $n$ starts from $1$.

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