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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.

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2 Answers 2

up vote 4 down vote accepted

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

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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 $p=3/2$.

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