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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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Do you know that for $|w|<1$ we have $\sum_{n=0}^\infty w^n=\frac{1}{1-w}$? – Dennis Gulko Nov 20 '12 at 0:35
@DennisGulko yes and I can potentially see a use for that here! Would I use a substitution? – Becky Nov 20 '12 at 0:53
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. – 000 Nov 20 '12 at 1:08

3 Answers

up vote 2 down vote accepted

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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Okay thank you, this is brilliant. So it's rather simple after all. All I do now is substitute back my z/5 and I'll have my answer. And I'm guessing the reason 0 < z < 5 is something to do with the 5 on the bottom, so I can have up to 5 on the top so it's less than 1 overall. – Becky Nov 20 '12 at 1:19
Oh, yeah. The range $0<|z|<5$ implies $0<|u|<1$, which 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. :P – 000 Nov 20 '12 at 1:21
up vote 3 down vote

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

Does this make it a bit more palatable?

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Thank you, can you possibly explain what it means by calculate a series, I've never come across it before. I'm guessing I'll transfer z into x + iy and get the answer in x and y's. – Becky Nov 20 '12 at 0:49
1  
Becky, you need to learn about geometric series. Begin here: en.wikipedia.org/wiki/Geometric_series – ncmathsadist Nov 20 '12 at 0:54
Oh! I know this is a geometric series and I know a bit about them, but I've never seen a question like this before where it's use the words calculate! So, could I substitute x and y's for z and use the limit of the standard geometric series 1/1-x? – Becky Nov 20 '12 at 0:59
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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. – Patrick Da Silva Nov 20 '12 at 1:13
up vote 1 down vote

HINT:

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

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Thank you, can you possibly explain what it means by calculate a series, I've never come across it before. I'm guessing I'll transfer z into x + iy and get the answer in x and y's. – Becky Nov 20 '12 at 0:35
@Becky: No need: just sum the geometric series $$\sum_{n\ge 0}\left(\frac{z}5\right)^n$$ and multiply by $\dfrac1{5z^2}$, and you’re done. – Brian M. Scott Nov 20 '12 at 6:00
I got 1/(5z^2 - z^3) correct? – Becky Nov 20 '12 at 22:21
@Becky: Yes, that looks fine. – Brian M. Scott Nov 20 '12 at 22:32

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