in a p-Series does the numerator have to equal 1, or can it be other constants?

ie does sum 6/(n^3) converge?

Thank You.

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- Dec 1st 2006, 12:45 PMredierRules for p-Series Convergence
in a p-Series does the numerator have to equal 1, or can it be other constants?

ie does sum 6/(n^3) converge?

Thank You. - Dec 1st 2006, 01:11 PMgalactus
You can take the 6 outside the sum:

$\displaystyle 6\sum_{n=1}^{\infty}\frac{1}{n^{3}}$

$\displaystyle \int_{1}^{+\infty}\frac{1}{x^{p}}=\lim_{l\to\infty }x^{-p}dx=\lim_{l\to\infty}\frac{x^{1-p}}{1-p}=\lim_{l\to\infty}\left[\frac{l^{1-p}}{1-p}-\frac{1}{1-p}\right]$

If p<1, then 1-p<0, so $\displaystyle l^{1-p}\rightarrow{0}$ as $\displaystyle l\rightarrow{+\infty}$, therefore,

the integral converges(it's value is $\displaystyle \frac{-1}{1-p}$) and consequently the series converges.

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{3}}=1+\frac{1}{8}+\ frac{1}{27}+\frac{1}{64}+\frac{1}{125}+.........=\ sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k+1)^{3}}=\frac{{\pi}^{3}}{32}$

This was one of Euler's answers. Unfortunately, it led to nowhere.

What do we know today about $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{3}}$?.

*Little.*

Maybe someone else can lend some insight. - Dec 1st 2006, 01:48 PMCaptainBlack
- Dec 2nd 2006, 12:14 PMredier
So when I justify my work should I use the integral test or say it's a p-Series. Because no matter what the constant is in the numerator it will always converge, (as long as the p > 1) right?

I already used the integral test to justify it, but I was wondering if that would be required. It would be a lot less writing to just say p-Series and be done with it. I guess it would also depend on the teacher too, but I was wondering you opinions on the matter, and what you would make your students do if you were a teacher. - Dec 2nd 2006, 01:24 PMCaptainBlack
It will depend on what you are allowed to assume, so if the convergence of

p-series has been established in an earlier part of your course, then usually

you can assume it (I would also assume convergence if I had proven it in an

earlier part of the same assignment - referring back to where it was established).

RonL - Dec 2nd 2006, 01:49 PMThePerfectHacker
Whoa! That is absolutely not true!

Euler was succesful in evaluating the zeta function for**even**numbers. This is odd.

It was only recent that a mathemation was succesful in finding the exact sum of $\displaystyle \zeta (3)$ (which is what you did). But the problem of $\displaystyle \zeta (2n-1),n\geq 1$ is still open. - Dec 2nd 2006, 02:06 PMgalactus
I know, PH, That's why I said it lead to nowhere. Maybe I should've elaborated more.