Infinite series convergence test

Question

Test the convergence of the following series:

$${\sqrt{n+1}-1\over (n+2)^3 -1} +... \infty$$

(This is a problem I got on my test today, I constructed a similar series without the -1 part and showed that since the quotient of the nth (n tending to infinity) term of the constructed series and the nth term of this series is non-zero, they behave alike and converge as tested by p series test, p = 5/2).

Answer 1

Since $$ \begin{align} \frac{\sqrt{n+1}-1}{(n+2)^3-1} &=\frac{1}{n^{5/2}}\frac{\sqrt{1+1/n}-\sqrt{1/n}}{(1+2/n)^3-1/n^3}\\ &\sim\frac1{n^{5/2}} \end{align} $$ your method looks fine.

Answer 2

\begin{align} \frac{\sqrt{n+1}-1}{(n+2)^3-1}&=\frac{(\sqrt{n+1}-1)(\sqrt{n+1}+1)}{((n+2)^3-1)(\sqrt{n+1}+1)}\\ &=\frac{n}{((n+2)^3-1)(\sqrt{n+1}+1)}\\ &<\frac{n}{n^3}\\ &=\frac{1}{n^2} \end{align} Using the p-test, $\frac{1}{n^2}$ converges, and since $0<\frac{n}{((n+2)^3-1)(\sqrt{n+1}+1)}<\frac{1}{n^2}$, using the comparison test, the original series must also converge.

Answer 3

It is clear that $\frac {\sqrt{n+1}-1}{(n+2)^3-1}\sim \frac{1}{n^\frac{5}{2}}$ as ${n\to\infty}$ . Since the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges so that the series must be converge