How do I compute the sum $$\sum_{k=0}^{\infty} (-1)^k (k+1) r^k$$ for $r \in [0,1)$? I know that the answer is $\frac{1}{(1+r)^2}$, but have no idea how to get there or how to even start. I know that $\sum_0^{\infty} (-1)^kr^k = 1/(1+r),$ but what to do with the $(k+1)$ I do not know.
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your sum is $$ (1+r)^{-2}+(1+r)^{-1} $$ – Jose Garcia May 27 '17 at 08:45
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1"I know that $\sum\limits_0^\infty(-1)^kr^k = 1/(1+r)$" Then differentiate both sides and rejoice... – Did May 27 '17 at 08:47
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1@JoseGarcia Checking the value at $r=0$ would help to avoid such statements... – Did May 27 '17 at 08:48
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Is there something wrong with my sum, I am not sure what you mean. @JoseGarcia – bwen95 May 27 '17 at 08:53
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Alright, I will try that, thank you. @Did – bwen95 May 27 '17 at 08:53
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See https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot – lab bhattacharjee May 27 '17 at 10:05
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@labbhattacharjee I'm dubious about the relevance of this link. – zwim May 27 '17 at 11:02
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@zwim, Compare $$\sum_{k=0}^{\infty} (-1)^k (k+1) r^k=1-2r+3r^2-4r^3+\cdots$$ with https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot/746975#746975 – lab bhattacharjee May 27 '17 at 11:35
2 Answers
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When you see a group like $(k+1)r^k$ you should think $(r^{k+1})'$
Thus for $|r|<1$ we have
$\displaystyle\sum\limits_{k=0}^{\infty}(-1)^k(k+1)r^k=-\bigg(\sum\limits_{k=0}^{\infty}(-1)^{k+1}r^{k+1}\bigg)'=-\bigg(\sum\limits_{k=1}^{\infty}(-r)^{k}\bigg)'=-\bigg(\frac{-r}{1+r}\bigg)'=\frac{1}{1+r}-\frac{r}{(1+r)^2}=\frac 1{(1+r)^2}$
zwim
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$S=\sum_{k=0}^{\infty} (-1)^k (k+1) r^k = 1-2r+3r^2-4r^3+\cdots+(-1)^k(k+1)r^k+\cdots$
$S^*=\int Sdr= r-r^2+r^3-r^4+\cdots+(-1)^kr^{k+1}+\cdots+C=\frac{r}{1+r}+C=1-\frac{1}{1+r}+C$
$S=(S^*)'=\frac{1}{(1+r)^2}$.
farruhota
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