Chapter 10.4, Problem 26E

To determine

To find: Interval of convergence and sum as a function of x for the given series.

Answer to Problem 26E

Interval of convergence is $\frac{1}{e}<x<e$ and sum is $S=\frac{1}{1-\ln x}$

Explanation of Solution

Given:

The series given is,

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(\ln x\right)}^n}$

Calculation:

A series is convergence when the limit of partial sum tends to infinity i.e. $n\to \infty$ , otherwise it is divergence.

Also, common ratio should be less than one i.e.

  $\left|r\right|<1$ where r is common ratio.

Now, the given series is

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(\ln x\right)}^n}$

Here, the above series is geometric series having common ratio $\ln x$

So,

  $\begin{array}{l}\left|\ln x\right|<1\\ \Rightarrow \frac{1}{e}<x<e\end{array}$

i.e. the interval of convergence is $\frac{1}{e}<x<e$

Now, determining the sum of the geometric series.

Here in $S={\displaystyle \sum _{n=0}^{\infty }{\left(\ln x\right)}^n}$ , the initial term is $a=1$ and common ratio $r=\ln x$

So, sum is

  $\begin{array}{l}S=\frac{a}{1-r}=\frac{1}{1-\ln x}=\frac{1}{1-\ln x}\\ \Rightarrow S=\frac{1}{1-\ln x}\end{array}$

Therefore, interval of convergence is $\frac{1}{e}<x<e$ and sum is $S=\frac{1}{1-\ln x}$