# Thread: Finding Value of P in Infinite Series

1. ## Finding Value of P in Infinite Series

$\sum_{n=1}^{\infty} \frac{4n^6}{3n^8+8}^{p}$
Find all values of p for which the infinite series converges.

Can any give me any advice or help on how to solve this problem? I have no idea where to begin.

2. ## Re: Finding Value of P in Infinite Series

Originally Posted by Beevo
$\sum_{n=1}^{\infty} \frac{4n^6}{3n^8+8}^{p}$
Find all values of p for which the infinite series converges.
Can any give me any advice or help on how to solve this problem? I have no idea where to begin.
I think that you mean $\sum_{n=1}^{\infty}\left( \frac{4n^6}{3n^8+8}\right)^{p}$

If so, then note that $\left( \frac{4n^6}{3n^8+8}\right)^{p}\le \left( \frac{4}{3n^2}\right)^{p}$

For that p-series what is the answer?

3. ## Re: Finding Value of P in Infinite Series

Let $\sum_{n=1}^{\infty} (\frac{4n^6}{3n^8+8})^{p} = \sum_{n=1}^{\infty} s_n$ where $s_n = (\frac{1}{n^2} \frac{4}{3+\frac{8}{n^8}})^{p}$ by Pringsheim Theorem we have the necessary condition that $\lim_{n \rightarrow \infty} n.s_n \rightarrow 0 \implies \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} (\frac{4}{3+\frac{8}{n^8}})^{p} =$ $\lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} \lim_{n \rightarrow \infty} (\frac{4}{3+\frac{8}{n^8}})^{p} = (\frac{4}{3})^p \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}}$ this limit $\rightarrow 0$ iff $2p-1 > 0 \implies p > \frac{1}{2}$.

4. ## Re: Finding Value of P in Infinite Series

Originally Posted by kalyanram
Let $\sum_{n=1}^{\infty} (\frac{4n^6}{3n^8+8})^{p} = \sum_{n=1}^{\infty} s_n$ where $s_n = (\frac{1}{n^2} \frac{4}{3+\frac{8}{n^8}})^{p}$ by Pringsheim Theorem we have the necessary condition that $\lim_{n \rightarrow \infty} n.s_n \rightarrow 0 \implies \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} (\frac{4}{3+\frac{8}{n^8}})^{p} =$ $\lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}} \lim_{n \rightarrow \infty} (\frac{4}{3+\frac{8}{n^8}})^{p} = (\frac{4}{3})^p \lim_{n \rightarrow \infty} \frac{1}{n^{2p-1}}$ this limit $\rightarrow 0$ iff $2p-1 > 1 \implies p>1 \$. We need $p>1$.
@kalyanram.
Why did you spoil the question for the student?
Do you want the student to learn? The student will never learn if you simply hand the complete solution.

5. ## Re: Finding Value of P in Infinite Series

Originally Posted by Plato
@kalyanram.
Why did you spoil the question for the student?
Do you want the student to learn? The student will never learn if you simply hand the complete solution.
I was typesetting in latex when you replied to the post. I was not aware of your post; wouldn't have simplified it if in that case.
~Kalyan.

6. ## Re: Finding Value of P in Infinite Series

Thank you for both the responses, but yeah, I was trying to figure out the problem on my own and the solution kind of spoiled it.

I don't why, but I am struggling to grasp the concept of series (especially the comparison tests). I felt Cal I was a breeze, but but Cal II is a different story so far. I am putting in at least 2-3 hours every other day of studying, but still having trouble keeping up. This is frustrating.