Question

Test for convergence:$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\frac{\sqrt{\mathbf{n}\mathbf{-}\mathbf{1}}}{{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$

Short answer

The limit is finite and the series converges.

Step-by-step solution

Concept used to prove that the series converges

Let $0⩽a_n⩽b_n$for all $n$.

Ifrole="math" localid="1664276577803" $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{b}}_{\mathbf{n}}$converges, then$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}$is also converges.

If$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}$diverges, then$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{b}}_{\mathbf{n}}$is also diverges.

Calculation to show that the series converges

Let the series $a_n=\sum _{n=2}^{\infty }\frac{\sqrt{n-1}}{{(n+1)}^2-1}$.

Consider the series $\sum _{n=2}^{\infty }\frac{\sqrt{n}}{n^2}=\sum _{n=2}^{\infty }\frac{1}{n^{\frac{3}{2}}}$.

Let $b_n=\frac{1}{n^{\frac{3}{2}}}$.

The series$\sum b_n=\sum \frac{1}{n^{\frac{3}{2}}}$is convergent series by integral test.

Take $li{m}_{n\to \infty }\frac{a_n}{b_n}$.

Simplify the above expression.

$$\begin{gathered} li{m}_{n\to \infty }\frac{a_n}{b_n}=li{m}_{n\to \infty }\left(\frac{\sqrt{n-1}}{{(n+1)}^2-1}÷\frac{1}{n^{\frac{3}{2}}}\right) \\ =li{m}_{n\to \infty }\left(\frac{\sqrt{n-1}}{{(n+1)}^2-1}·n^{\frac{3}{2}}\right) \\ =li{m}_{n\to \infty }\left(\frac{\sqrt{n}\sqrt{1-\frac{1}{n}}}{n^2\left({\left(1+\frac{1}{n}\right)}^2-\frac{1}{n^2}\right)}·n^{\frac{3}{2}}\right) \end{gathered}$$

Solve the limit.

width="244">limn→∞anbn=1-1w1+1w2-1(s)2

Hence, the limit is finite and the series converges.