Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove that series are divergent/convergent:

$\displaystyle\sum_{n=2}^{\infty}(n\sqrt{n}-\sqrt{n^3-1})$

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|improve this question
add comment

2 Answers

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|improve this answer
add comment

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|improve this answer
add comment

Your Answer

 
discard

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.