Chapter 8.1, Problem 19E

To determine

To find: To determine whether the sequence converges or diverges and find limit of converges.

Answer to Problem 19E

The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=0}$

Explanation of Solution

Given information: $a_n=\frac{{\left(-1\right)}^{n-1}\cdot n}{n^2+1}$

Solution:

To determine converges or diverges we need to check the limit of the sequence of the partial sums.

  $\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^{n-1}\cdot n}{n^2+1}$

Divide numerator and denominator by $\text{'}{n}^2\text{'}.$ we will get,

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{\frac{{\left(-1\right)}^{n-1}}{n}}{1+\frac{1}{n^2}}\\ =\frac{\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^{n-1}}{n}}{\underset{n\to \infty }{\lim }\left(1+\frac{1}{n^2}\right)}\\ =\frac{0}{1}\\ =0\end{array}$

  $\therefore$ Limit exists and is finite.

Hence, The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=0}$