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.
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.
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.
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
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.