Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove that series are divergent/convergent:


I tried using Limit comparison (with $1/n$), Root and Ratio tests, but they gave no result. With integral test I was left with $\displaystyle\int_{2}^{\infty}\sqrt{n^3-1}\mathrm dn$.

Any suggestions? Should I use comparison test, with different series? Thanks.

share|cite|improve this question
up vote 4 down vote accepted

Hint: $$\sqrt{n^3}-\sqrt{n^3-1} = \frac{1}{\sqrt{n^3}+\sqrt{n^3-1}} \sim \frac{1}{2n^{3/2}}.$$

share|cite|improve this answer

Use this good fact that if $\lim_{n\to\infty}~n^pu_n=A$ and $p>1$ and $A$ is finite then $\sum u_n$ converges. This and the good hint of @njguliyev made a solution completed. Here $p=3/2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.