1. ## convergence and divergence

Consider
∞ ∑ (3k+5/k^2-2k)^pk=3
for each p ∈ R. show this series - converges if p > 1
- diverges if p < or = to 1

Hint: Determine the known series whose terms past the second give an approximate match for the terms of this series. This series is suitable (almost) for using the comp[arison test. Seperate comparisons with it, or closely related series, are needed to establish convergence or divergence of the series according to p. You will need to establish inequalities, based on approximations, to apply the comparison tests.

2. $\displaystyle \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p$

note:

$\displaystyle \frac{1}{k} < \frac{3}{k} < \frac{3}{k} + \frac{5}{k^2} = \frac{3k+5}{k^2} < \frac{3k+5}{k^2-2k}$

so, if $\displaystyle p=1$
$\displaystyle \sum_{k=3}^{\infty} \frac{1}{k}$ diverges..

if $\displaystyle 0<p<1$,
$\displaystyle \sum_{k=3}^{\infty} \left(\frac{1}{k}\right)^p$ also diverges by p-series test.

also, if $\displaystyle p\leq0$,
$\displaystyle \sum_{k=3}^{\infty} \left(\frac{1}{k}\right)^p = \sum_{k=3}^{\infty} k^{-p}$ also diverges..

hence, $\displaystyle \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p$ diverges for $\displaystyle p\leq1$ by comparison test.

3. $\displaystyle \frac{3k+5}{k^2-2k} < \frac{3k+5}{-2k} = -\frac{3}{2}-\frac{5}{2k} < -\frac{5}{2k}$

so, $\displaystyle \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p < \sum_{k=3}^{\infty} \left(-\frac{5}{2k}\right)^p$

and $\displaystyle \sum_{k=3}^{\infty} \left(-\frac{5}{2k}\right)^p$ converges for $\displaystyle p>1$.

therefore $\displaystyle \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p$ converges for $\displaystyle p>1$.