Convergence/Divergence Question

**ineedhelplz** · August 7th 2011, 11:09 PM

I have spent over 5 hours trying to solve this question to no avail.

**Bacterius** · August 7th 2011, 11:20 PM

What have you tried? What can you say about the inequality $3k + 5 < k^2 - 2k$ when $k \geqslant 3$ . Once you've established that, what bounds can you put on $\frac{3k + 5}{k^2 - 2k}$ ? Hence, how does the value of $p$ 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 $p$ ?

**CaptainBlack** · August 8th 2011, 01:25 AM

Originally Posted by ineedhelplz

I have spent over 5 hours trying to solve this question to no avail.

Since for large $k$ :

$\frac{3k+5}{k^2-2k} \sim \frac{3}{k}$

if show that there is a natural $N_0$ such that for all $k>N_0$ , there exist positive constanta $A$ and $B$ such that:

$A \frac{3}{k}< \frac{3k+5}{k^2-2k}< B \frac{3}{k}$

convergence of the given series is reduced to convergence of:

$\sum_{k=3}^{\infty} \frac{1}{k^p}$

CB

**ineedhelplz** · August 9th 2011, 03:59 PM

Originally Posted by CaptainBlack

Since for large $k$ :

$\frac{3k+5}{k^2-2k} \sim \frac{3}{k}$

if show that there is a natural $N_0$ such that for all $k>N_0$ , there exist positive constanta $A$ and $B$ such that:

$A \frac{3}{k}< \frac{3k+5}{k^2-2k}< B \frac{3}{k}$

convergence of the given series is reduced to convergence of:

$\sum_{k=3}^{\infty} \frac{1}{k^p}$

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:
$\frac{3k+5}{k^2-2k} \sim \frac{3}{k}$

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...

**Also sprach Zarathustra** · August 9th 2011, 04:37 PM

Originally Posted by ineedhelplz

I have spent over 5 hours trying to solve this question to no avail.

With limit comparison test.

$A$ is series with $a_k=(\frac{3k+5}{k^2-2k})^p$ , and $B$ is a series with $b_k=(\frac{3}{k})^p$

$\lim_{k\to\infty} \frac{a_k}{b_k}=\lim_{k\to\infty}(\frac{\frac{3k+5 }{k^2-2k}}{\frac{3}{k}})^p=1^p=1$

Hence $A$ and $B$ converges or diverges together.

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