Chapter 10.1, Problem 62E

(a)

To determine

The value of the limit.

Answer to Problem 62E

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{k=n}{a}_k}$ converges to S.

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Formula used:

  ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Calculation:

Given function is,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{k=n}{a}_k}$ On applying the given limit

  ${\displaystyle \sum _{k=1}^{k=\infty }{a}_k}$ Since ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Therefore,

  ${\displaystyle \sum _{k=1}^{k=\infty }{a}_k}={\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Thus,

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{k=n}{a}_k}$ converges to S.

Conclusion:

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{k=n}{a}_k}$ converges to S.

(b)

To determine

To show: $S_{n-1}+a_n=S_n$

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Formula used:

  $S=\frac{a-a{r}^n}{1-r}$

Calculation:

The sum of any geometric series $S=\frac{a-a{r}^n}{1-r}$ where, a is initial term, r is the common ratio and n is the number of terms. Therefore, sum of n partial terms is $S_n=\frac{a-a{r}^n}{1-r}$ sum of n-1 partial terms is $S_{n-1}=\frac{a-a{r}^{n-1}}{1-r}{n}^{th}$ term of the series, $a_n=a{r}^{n-1}$

Therefore,

  $\begin{array}{l}{S}_{n-1}+a_n=\frac{a-a{r}^{n-1}}{1-r}+a{r}^{n-1}\\ =\frac{a-a{r}^{n-1}+a{r}^{n-1}\left(1-r\right)}{1-r}\\ =\frac{a-a{r}^{n-1}+a{r}^{n-1}-a{r}^{n-1}\gamma }{1-r}\\ =\frac{a-a{r}^n}{1-r}\\ S_{n-1}+a_n=S_n\end{array}$

Conclusion:

  $S_{n-1}+a_n=S_n$

(c)

To determine

To show: The value of $\underset{n\to \infty }{\lim }{a}_n=0$ .

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

Formula used:

  $a_n=a{r}^{n-1}$

Calculation:

Since, ${\displaystyle \sum _{k=1}^{\infty }{a}_k}=S$

If geometric series of infinite terms converges then, $\left|r\right|<1$ . Where r is the common ratio of the series n^thterm of the series, $a_n=a{r}^{n-1}$ where a is initial term, r is the common ratio and n is the number of terms. Therefore

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }a{r}^{n-1}\\ =a\underset{n\to \infty }{\lim }{r}^{n-1}\\ =a.0\\ \underset{n\to \infty }{\lim }{a}_n=0\end{array}$

(number x such that $\left|x\right|<1$ , then $x^{\infty }\to 0$

Thus,

  $\underset{n\to \infty }{\lim }{a}_n=0$

Conclusion:

  $\underset{n\to \infty }{\lim }{a}_n=0$