Question

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

Short answer

The series$\sum _{n=2}^{\infty }\frac{{(n-1)}^2}{1+n^2}$ diverges.

Step-by-step solution

Concept used to prove that the series diverges

Conditions of convergence for the series $\sum _{n=1}^{\infty }{a}_n$are:

$li{m}_{n\to \infty }{a}_n=0$

Calculation used to show that the series ∑n=2∞(n-1)21+n2  diverges

${\text{n}}^{\text{th}}$term of the series is,$a_n=\frac{{(n-1)}^2}{1+n^2}$

Apply limit in $n^{\text{th}}$terms of the series as follows:

$$\begin{gathered} li{m}_{n\to \infty }{a}_n=li{m}_{n\to \infty }\frac{{(n-1)}^2}{1+n^2} \\ li{m}_{n\to \infty }{a}_n=li{m}_{n\to \infty }\frac{n^2{\left(1-\frac{1}{n}\right)}^2}{n^2\left(\frac{1}{n^2}+1\right)} \\ li{m}_{n\to \infty }{a}_n=li{m}_{n\to \infty }\frac{{\left(1-\frac{1}{w}\right)}^2}{\left(\frac{1}{a^2}+1\right)} \\ li{m}_{n\to \infty }{a}_n=\frac{1}{1} \\ li{m}_{n\to \infty }{a}_n=1 \end{gathered}$$

Thus, this does not prove the second condition of the convergence as $li{m}_{n\to \infty }{a}_n\ne 0$.

Hence, the series diverges.