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I want to find the sum of a given power series:

$$\sum_{n=0}^\infty(n+4)x^{n-3}$$

I'm trying to find the sum through slow integration or differentiation of a series. But, unfortunately, I can't find the right combination for this, although it's probably a simple one. I know there are a bunch of other ways to find the sum of a power series, but I'm interested in solving this example through slow integration/differentiation.

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    $\begingroup$ $n-3=(n+4) -7$. $\endgroup$
    – cosmo5
    Oct 2, 2020 at 14:48
  • $\begingroup$ the first term is $4x^{-3}$. is this right? Not quite a power series then. $\endgroup$
    – lhf
    Oct 2, 2020 at 14:49

2 Answers 2

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\begin{align} \sum_{n=0}^\infty(n+4)x^{n-3}&=x^{-6}\sum_{n=0}^\infty(n+4)x^{n+3}\\ &=x^{-6}\frac d{dx}\left(\sum_{n=0}^\infty x^{n+4}\right)\\ &=x^{-6}\frac d{dx}\left(\frac{x^4}{1-x}\right).\\ \end{align}

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Alternatively $$\sum_{n=0}^{\infty}(n+4)x^{n-3}$$ $$=x^{-3}\sum_{n=0}^{\infty}(n+4)x^n$$ $$=x^{-3}\big[x\sum_{n=0}^{\infty}nx^{n-1}+4\sum_{n=0}^{\infty}x^{n}\big]$$ $$=x^{-2}\frac{d}{dx}\big(\frac{1}{1-x}\big)+\frac{4}{x^{3}(1-x)}$$ $$=\frac{1}{x^2(1-x)^2}+\frac{4(1-x)}{x^{3}(1-x)^2}$$ $$=\frac{1}{x^3(x-1)^2}\big[x+4(1-x)\big]=\frac{4-3x}{x^3(1-x)^2}$$ when $|x|<1$.

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