# Thread: Testing Series for Convergence/Divergence

1. ## Testing Series for Convergence/Divergence

Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
I dont see how the ratio test could work here.

any pointers? thanks!

Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
I dont see how the ratio test could work here.

any pointers? thanks!
Try this

$\displaystyle \sum_{n=1}^{\infty} \frac{n \ln n}{(n+8)^3} \le \sum_{n=1}^{\infty} \frac{n \ln n}{n^3} = \sum_{n=1}^{\infty} \frac{\ln n}{n^2}$ then use the integral test on the last series.

Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
I dont see how the ratio test could work here.

any pointers? thanks!
I kind of have the same approach as Danny but not as efficient as his.

Use the limit test

$\displaystyle a_n = \frac{n\ln{n}}{(n+8)^3}$

$\displaystyle b_n = \frac{\ln{n}}{n^2}$

so

$\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n} = 1$

so now if $\displaystyle b_n$ converges so does $\displaystyle a_n$

Use the integral test to see if $\displaystyle b_n$ converges

$\displaystyle u= \ln{n}$

$\displaystyle du = \frac{dn}{n}$

$\displaystyle dv = n^{-2}dn$

$\displaystyle v = -n^{-1}$

$\displaystyle \int^{\infty}_{1} = -\frac{\ln{n}}{n} +\int \frac{dn}{n^2}$

$\displaystyle -\frac{\ln{n}}{n} - \frac{1}{n}\bigg|^{\infty}_{1}$

which converges to a number using L' Hopital.

so since $\displaystyle b_n$ converges so does $\displaystyle a_n$