# 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!

#### Jester

MHF Helper
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.

#### 11rdc11

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 $$\displaystyle a_n$$