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How would one compute $ \sum_{n=0}^\infty\frac{z^{n-2}}{5^{n+1}} $ where $0\lt|z|\lt5$?

I have literally no idea where to start, all I know is that the answer will not have summations. Any help would be appreciated!

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    Yep. That's precisely it. You don't have to consider it a substitution; but, if it helps you understand the manipulation, then go for it.2012-11-20

3 Answers 3

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HINT:

$\frac{z^{n-2}}{5^{n+1}}=\frac1{5z^2}\left(\frac{z}5\right)^n$

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    @Becky: Yes, that looks fine.2012-11-20
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$\sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n$

Does this make it a bit more palatable?

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    @Becky : you really don't need to go back in terms of $z = x+ iy$. An answer in terms of $z$ is perfectly suitable. If you want to actually "compute the number, given say $z = 1 + i$", then find the "nice-looking expression not involving series" for $z$, then substitute in and make the computations.2012-11-20
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As said above,

$ \sum_{n=0}^\infty {z^{n-2}\over 5^{n+1}}= {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n. $

To solve this, it may aid you to make the substitution $u=\frac{z}{5}$. Then, $ {1\over 5z^2}\sum_{n=0}^\infty\left({z\over 5}\right)^n=\frac{1}{5z^2}\sum_{n=0}^{\infty}u^n. $

The sum is geometric in $u$; thus, you apply the geometric series formula.

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    Oh, yeah. The range 0<|z|<5 implies 0<|u|<1, w$h$ich satisfies the condition of a geometric series converging to a finite value. If you'd like, feel free to click the little checkmark and vote this answer up. I'd like to hit 1k tonight. :P2012-11-20