I have spent over 5 hours trying to solve this question to no avail.
What have you tried? What can you say about the inequality $\displaystyle 3k + 5 < k^2 - 2k$ when $\displaystyle k \geqslant 3$. Once you've established that, what bounds can you put on $\displaystyle \frac{3k + 5}{k^2 - 2k}$? Hence, how does the value of $\displaystyle p$ affect the partial sums? Use either the comparison test (which I think you are meant to use) or another method to determine the convergence/divergence of the sum in relation to $\displaystyle p$?
Since for large $\displaystyle k$:
$\displaystyle \frac{3k+5}{k^2-2k} \sim \frac{3}{k}$
if show that there is a natural $\displaystyle N_0$ such that for all $\displaystyle k>N_0$, there exist positive constanta $\displaystyle A$ and $\displaystyle B$ such that:
$\displaystyle A \frac{3}{k}< \frac{3k+5}{k^2-2k}< B \frac{3}{k}$
convergence of the given series is reduced to convergence of:
$\displaystyle \sum_{k=3}^{\infty} \frac{1}{k^p}$
CB
Thank you for your reply.
I'm unsure as to the next step I'm supposed to take. I presume that the first step "Determine the known series whose terms past the second give an approximate match for the terms of this series." is solved by what you wrote here:
$\displaystyle \frac{3k+5}{k^2-2k} \sim \frac{3}{k}$
however, I am unsure how to do the second step "establish inequalities, based on approximations (as below), to apply the comparison test". I know how to apply the comparison test, but don't know how to use the approximations...
With limit comparison test.
$\displaystyle A$ is series with $\displaystyle a_k=(\frac{3k+5}{k^2-2k})^p$ , and $\displaystyle B$ is a series with $\displaystyle b_k=(\frac{3}{k})^p$
$\displaystyle \lim_{k\to\infty} \frac{a_k}{b_k}=\lim_{k\to\infty}(\frac{\frac{3k+5 }{k^2-2k}}{\frac{3}{k}})^p=1^p=1$
Hence $\displaystyle A$ and $\displaystyle B$ converges or diverges together.