Determine whether $\sum_{n=0}^\infty\frac n{n^2\sqrt n+1}$ converges

Question

$$\sum_{n=0}^\infty\frac n{n^2\sqrt n+1}$$ I tried to use the ratio test but my answer is 1 and that gives no information about the answer. I have also used the root test and the given answer is 1. How could I determine the convergence or divergence of this series? I have also tried to use the integral test and the answer is too long and to difficult.

Answer 1

HINT

Note that

$$ \frac{n}{(n^2 {\sqrt n} +1)}\sim \frac1{n\sqrt n}$$

then use limit comparison test with $\sum \frac1{n\sqrt n}$ to show convergence.

Answer 2

$$ \sum_n \frac{n}{n^{5/2}+1} \le \sum_n \frac{1}{n^{3/2}} < \infty. $$

Answer 3

$$\sum_{n=0}^\infty\frac n{n^2\sqrt n+1}=\sum_{n=1}^\infty\frac n{n^2\sqrt n+1}\le\sum_{n=1}^\infty\frac n{n^2\sqrt n}=\sum_{n=1}^\infty\frac1{n^{3/2}}=\zeta(3/2)^*<\infty$$

*The series representation $\zeta(x)=\sum_{n=1}^\infty n^{-x}$ is convergent for all $x>1$.