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 \(\displaystyle a_n\)