real infinite series

Question

Let $p>1$. Prove that $$\sum^{\infty}_{n=1} \frac {\sqrt[p] {n^{p-1}}}{(n+1)n} <\infty.$$

I got that the numerator is less than $n$ so that I want to find something that is lower than the denominator. Thus I could use the comparison theorem and see that converges. Am I on the right track?

Answer 1

HINT

Note that for $n$ large

$$ \frac {\sqrt[p] {n^{p-1}}}{(n+1)n}\sim \frac{1}{n^{2-\frac{p-1}p}}=\frac{1}{n^{1+\frac1p}}$$

with

$$\alpha=1+\frac1p>1$$

for $p>1$, then we can apply the limit comparison test with $\sum \frac{1}{n^{1+\frac1p}}$.

Answer 2

The $n$th term equals

$$\frac{n^{1-1/p}}{n(n+1)} = \frac{1}{n^{1/p}(n+1)} < \frac{1}{n^{1/p}n}=\frac{1}{n^{1+1/p}}.$$

Apply the comparison test.