Chapter 10, Problem 45RE

To determine

To check: If the series converges absolutely, converges conditionally or diverges.

Answer to Problem 45RE

Series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{\ln \left(n+1\right)}}$ converges conditionally

Explanation of Solution

Given:

The integral given is, ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{\ln \left(n+1\right)}}$

Calculation:

Compare the terms of ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{\ln \left(n+1\right)}}$ with the terms of ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$

By limit comparison test,

  $\underset{n\to \infty }{\lim }\frac{\frac{1}{\ln (n+1)}}{\frac{1}{n}}=\underset{n\to \infty }{\lim }\frac{n}{\ln \left(n+1\right)}=\underset{n\to \infty }{\lim }\frac{1}{1/\left(n+1\right)}=\infty$

Thus, the series doesn’t converge absolutely.

Now, use the Alternating Series test to

Verify the Conditional convergence. $u_n=-\frac{1}{\ln \left(n+1\right)}>0$ ….(1)

Asn+ 1 >n i.e. ln( n +1)>ln n

  $\Rightarrow -\frac{1}{\ln (n+1)}<\frac{1}{\ln n}$ ….. (2)

i.e. $u_n$ is decreasing.

  $\underset{n\to \infty }{\lim }{u}_n$ = $\underset{n\to \infty }{\lim }\frac{1}{\ln n}$

  $\Rightarrow \underset{n\to \infty }{\lim }{u}_n=0$ …….(3)

From (1),(2) and (3), the Alternate test condition satisfies.

Therefore, ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{\ln \left(n+1\right)}}$ converges conditionally.