Chapter 10.4, Problem 25E

To determine

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

Answer to Problem 25E

Interval of convergence is $0<x<16$ and sum is $S=\frac{2}{4-\sqrt{x}}$

Explanation of Solution

Given:

The series given is,

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{\sqrt{x}}{2}-1\right)}}^n$

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

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{\sqrt{x}}{2}-1\right)}}^n$

Here, the above series is geometric series having common ratio $\left(\frac{\sqrt{x}}{2}-1\right)$

So,

  $\begin{array}{l}\left|\frac{\sqrt{x}}{2}-1\right|<1\\ \Rightarrow 0<x<16\end{array}$

i.e. the interval of convergence is $0<x<16$

Now, determining the sum of the geometric series.

Here in $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{\sqrt{x}}{2}-1\right)}}^n$ , the initial term is $a=1$ and common ratio $r=\left(\frac{\sqrt{x}}{2}-1\right)$

So, sum is

  $\begin{array}{l}S=\frac{a}{1-r}=\frac{1}{1-\left(\frac{\sqrt{x}}{2}-1\right)}=\frac{2}{4-\sqrt{x}}\\ \Rightarrow S=\frac{2}{4-\sqrt{x}}\end{array}$

Therefore, interval of convergence is $0<x<16$ and sum is $S=\frac{2}{4-\sqrt{x}}$