$((n-1)^{0.5})/(((n+1)^2)-1)$ Is the sum convergent?, why or why not?

Question

$$\frac{(n-1)^{0.5}}{(n+1)^2-1}$$

Sorry I dont know how to to do sub or superscripts. I would like a step by step method please, thanks.

Answer 1

It is convergent. To see this, use equivalents:

• $(n-1)^{1/2}\sim_\infty n^{1/2}$
• $(n+1)^2-1\sim_\infty n^2$

Hence $\;\dfrac{(n-1)^{1/2}}{(n+1)^2-1}\sim_\infty\dfrac{n^{1/2}}{n^2}=\dfrac1{n^{3/2}}$, which converges.