Chapter 8.4, Problem 39E

To determine

To calculate: Whether the given series is absolute convergent or not.

Answer to Problem 39E

The given series is absolute convergent.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }{\left(\frac{n^2+1}{2n^2+1}\right)}^n}$

Concept Used:

Root test: Whether the series is absolute convergent

  $\underset{n\to \infty }{\lim }\sqrt[n]{\left|a_n\right|}=L$

(i) If $L<1$ , then ${\displaystyle \sum a_n}$is absolute convergent.

(ii) If $L>1$ $(\text{or}L=\infty )$ , then ${\displaystyle \sum a_n}$is divergent.

(iii) If $L=1$, then the root test is inconclusive.

Calculation:

The series is ${\displaystyle \sum _{n=1}^{n=\infty }{\left(\frac{n^2+1}{2n^2+1}\right)}^n}$

By using the root test for this series to check it is absolute convergent.

  $\underset{n\to \infty }{\lim }\sqrt[n]{\left|a_n\right|}=L$

So, for the given series the $a_n$ is

  $a_n={\left(\frac{n^2+1}{2n^2+1}\right)}^n$

This term $a_n$ is positive for all value of $n\ge 1$

So,

  $a_n=\left|a_n\right|$

Then put the value of $a_n$ in the root test equation

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\sqrt[n]{\left|a_n\right|}\\ =\underset{n\to \infty }{\lim }\sqrt[n]{a_n}\\ =\underset{n\to \infty }{\lim }\sqrt[n]{{\left(\frac{n^2+1}{2n^2+1}\right)}^n}\\ =\underset{n\to \infty }{\lim }\left(\frac{n^2+1}{2n^2+1}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1+\frac{1}{n^2}}{2+\frac{1}{n^2}}\right)\\ =\left(\frac{1+\frac{1}{n^2}}{2+\frac{1}{n^2}}\right)\to \frac{1}{2}asn\to \infty \\ \end{array}$

So,

  $\begin{array}{l}L=\frac{1}{2}asn\to \infty \\ L=\frac{1}{2}<1\end{array}$

According to root test if $L<1$ then ${\displaystyle \sum a_n}$ is absolute convergent.

Conclusion:

The given series is absolute convergent.