Chapter 10.1, Problem 61E

(a)

To determine

The value of S in which t=1.

Answer to Problem 61E

The value is ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n}=2$

Explanation of Solution

Given information:

The given series is,

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{t}{1+t}\right)}}^n$

Formula used:

ratio is $<1.S=\frac{a}{1-r}$

Calculation:

Putting t=1 in the above expression,

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

The sum of the infinite geometric series whose ratio is $<1.S=\frac{a}{1-r}$ ’Where, a is initial term and r is the common ratio.

It is a geometric progression with initial term is 1 and common ratio is equal to $\frac{1}{2}$

Therefore, sum of the series is

  $\begin{array}{l}S=\frac{1}{\left(1-\frac{1}{2}\right)}\\ =2\\ {\displaystyle \sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n}=2\end{array}$

Conclusion:

The value is ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n}=2$

(b)

To determine

The value of t when S converges.

Answer to Problem 61E

  $S$ converges if $t>-\frac{1}{2}$

Explanation of Solution

Given information:

The given series is,

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{t}{1+t}\right)}}^n$

Formula used:

The geometric progression is used.

Calculation:

If the sequence of partial sum has a limit as $n\to \infty$ , the series converges.

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{t}{1+t}\right)}}^n$ It is a geometric progression with initial term is 1 and common ratio is equal to $\left(\frac{t}{1+t}\right)$ . The series converges when $\left|r\right|<1$ , that is $\left|\left(\frac{t}{1+t}\right)\right|<1\text{or}\left|t\right|<\left|1+t\right|\text{if }t<-1,$ that gives $-t<-\left(1+t\right)$ , which is always false if $-1\le t<0$ , that gives $-t<1+t$ , which is true when $t>-\frac{1}{2}$ . If $t\ge 0$ , that gives $t<1+t$ , this is always true. Therefore, $\left|\left(\frac{t}{1+t}\right)\right|<1,\text{if}t>-\frac{1}{2}$ . Thus, $S$ converges if $t>-\frac{1}{2}$

Conclusion:

  $S$ converges if $t>-\frac{1}{2}$

(c)

To determine

The value of t when S is greater than 10.

Answer to Problem 61E

The sum of the series is $S=\frac{1}{1-\left(\frac{t}{1+t}\right)}=1+t$ therefore, $S>10$ is $t>9$

Explanation of Solution

Given information:

The given series is,

  $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{t}{1+t}\right)}}^n$

Formula used:

The infinite geometric series is used.

Calculation:

For $t>-\frac{1}{2}$ , have $S={{\displaystyle \sum _{n=0}^{\infty }\left(\frac{t}{1+t}\right)}}^n$ . The sum of the infinite geometric series whose ratio is less than 1 $S=\frac{a}{1-r}$ where, a is initial term and r is the common ratio. It is a geometric progression with initial term is 1 and common ratio is equal to $\left(\frac{t}{1+t}\right)$ . Therefore, sum of the series is $S=\frac{1}{1-\left(\frac{t}{1+t}\right)}=1+t$ therefore, $S>10$ is $t>9$

Conclusion:

The sum of the series is $S=\frac{1}{1-\left(\frac{t}{1+t}\right)}=1+t$ therefore, $S>10$ is $t>9$