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