Convergence and divergence of $\sum_{n=2}^\infty {1\over{\sqrt[3]{n^2-1}}}$ and $\sum_{n=1}^\infty {1\over{\sqrt[3]{n^2+1}}}$

Question

I need to decide if these series are convergent or divergent, using the comparison test, the quotient test, the integral test, or the Leibniz Theorem:

$$\sum_{n=2}^\infty {1\over{\sqrt[3]{n^2-1}}}$$ $$\sum_{n=1}^\infty {1\over{\sqrt[3]{n^2+1}}}$$

For the first one, I think that when $n$ is big, $1\over{\sqrt[3]{n^2-1}}$ is almost equal to $1\over{\sqrt[3]{n^2}}$, right? I think I can use this to do a comparison test. $1\over{\sqrt[3]{n^2}}$ diverges because it is equal to $1\over{n^{2/3}}$ and ${2\over3}<1$? How do I solve this; is my reasoning correct? Thanks.

Answer 1

$1\over{\sqrt[3]{n^2-1}}$ >${1\over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.

Write $\sqrt[3]{n^2-1}=n^{2/3}\sqrt[3]{1-{1\over n^{2/3}}}$

$\sqrt[3]{n^2+1}=n^{2/3}\sqrt[3]{1+{1\over n^{2/3}}}$.

Answer 2

$\sqrt [3] {n^{2}-1} \leq n^{2/3}$ for all $n >1$. Hence the first series dominates $\sum \frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$\sqrt [3] {n^{2}+1} \leq 2n^{2/3}$].

Answer 3

Let $n \ge 2:$

$\dfrac{1}{2\sqrt[3]{n^2}} \lt \dfrac{1}{\sqrt[3]{n^2+n^2}} \lt$

$\dfrac{1}{\sqrt[3]{n^2+1}} \lt \dfrac{1}{\sqrt[3]{n^2-1}}.$

$(1/2)\sum \dfrac{1}{\sqrt[3]{n^2}}$ diverges.

Hence?