Answer to Question #143346 in Calculus for Moel Tariburu

Question

Determine whether or not the following series converges. ∑_(k=1)^∞▒8k/√(k^2+5)

Answer 1

$∑_{k=1}^∞\frac{8k}{\sqrt{(k^2+5)}}$

Keeping only the fastest growing terms in the numerator abd denominator, we have;

$∑_{k=1}^∞\frac{8k}{\sqrt{k^2}}=∑_{k=1}^∞\frac{8k}{k}=\\ ∑_{k=1}^∞8$

p = 0

This is a p-series with p $\le$ 0 so it is divergent.

We use the limit comparison test to compare the series.

$∑_{k=1}^∞ak = ∑_{k=1}^∞\frac{8k}{\sqrt{(k^2+5)}}$

$∑_{k=1}^∞bk = ∑_{k=1}^∞8$

$\begin{aligned} \therefore \dfrac{ak}{bk} &= \dfrac{\frac{8k}{\sqrt{(k^2+5)}}}{8}\\ \\ &= \dfrac{16k}{\sqrt{(k^2+5)}}\\ \\ &= \dfrac{\sqrt{256k^2}}{\sqrt{(k^2+5)}}\\ \\ &= \sqrt{\dfrac{256k^2}{(k^2+5)}}\\ \\ &= \sqrt{\dfrac{1}{1/256 + 5/k^2}}\\ \\ &\to 1= L \end{aligned}$

According to the limit comparison test, if 0<L<$\infin$, we can conclude that both series are divergent if one of them is.

$\therefore ∑_{k=1}^∞\frac{8k}{\sqrt{(k^2+5)}}$ is divergent