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There are multiple ways to derive $\sum_{x=1}^\infty xr^x = {r\over (r-1)^2}$ mentioned here How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? but none of them show the derivation by taking the limit of the partial sum formula. I'm having trouble with this.

Starting from the version of the formula Wolfram alpha gave: $$\sum_{x=1}^n xr^x = {\frac{(nr-n-1)r^{n+1}+r}{(1-r)^2}}$$

$$\lim_{n\rightarrow \infty}{\frac{(nr-n-1)r^{n+1}+r}{(1-r)^2}}$$

$$=\frac{1}{(1-r)^2}\bigg(\lim_{n\rightarrow \infty} (nr^{n+2}) - \lim_{n\rightarrow \infty} (nr^{n+1}) -r^{n+1} + r\bigg)$$

Using L'Hopital's rule and the fact that $|r|<1$

$$\lim_{n\rightarrow \infty} (nr^{n+2}) =- \lim_{n\rightarrow \infty} (nr^{n+1}) =0$$

So we have $$=\frac{1}{(1-r)^2}\big( -r^{n+1} + r\big)$$

which does not equal $\frac {r}{(r-1)^2}$. Where is my mistake?

xojfqa
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    In fact, $\sum_{x=1}^\infty r^x = \frac{r}{1-r}$, and not $\sum_{x=1}^n x r^x$. – GEdgar Oct 03 '22 at 21:30
  • @GEdgar Sorry, I forgot the square in the denominator. The equation is $\sum_{x=1}^\infty xr^x=\frac{r}{(r-1)^2}$, as confirmed by the proofs in the linked post. But my derivation from the limit is still incorrect – xojfqa Oct 03 '22 at 21:42

2 Answers2

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Don't forget that $\lim_{n\to\infty} r^{n+1}=0$ if $|r|<1$.

grand_chat
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$\displaystyle \sum_{x=1}^{\infty} xr^x=\frac{r}{(r-1)^2}.$

Well done! You've already proved this one.

$\displaystyle \sum_{x=1}^{\infty} xr^x=\lim_{n \to \infty}\frac{1}{(1-r)^2}\left(-r^{n+1}+r\right) = \frac{r}{(1-r)^2}$.

Adding my classic solution, too...

\begin{align} & \sum_{x=1}^{\infty} xr^x=S. \\ S & = 1r^1+2r^2+3r^3+4r^4+\cdots \\ rS & =0r^1+1r^2+2r^3+3r^4+\cdots \\ \therefore \; & (1-r)S=\sum_{x=1}^{\infty} r^x = \dfrac{r}{1-r}. \\ \therefore \; & S = \dfrac{r}{(r-1)^2}. \end{align}

RDK
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