Chapter 8.4, Problem 20E

To determine

To calculate:The sum of series correct to four decimal places.

Answer to Problem 20E

The sum of the series is $s=-0.2834$ correct to four decimal places.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{3^{n}n!}}$

Concept Used:

Alternating Series Estimation Theorem:

If $s={\displaystyle \sum {\left(-1\right)}^{n-1}{b}_n}$ is the sum of an alternating series that satisfies

  (i) $b_{n+1}\le b_n$

  (ii) $\underset{n\to \infty }{\lim }{b}_n=0$

Then the series is convergent and

  $\left|s-s_n\right|\le b_{n+1}$

Calculation:

The series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{3^{n}n!}}$

Firstly, observe that the series is convergent by alternating series test

Here,

  $b_n=\frac{1}{3^{n}n!}$

Then

  $b_{n+1}=\frac{1}{3^{n+1}\left(n+1\right)!}$

(i) $\begin{array}{l}{b}_{n+1}=\frac{1}{3^{n+1}\left(n+1\right)!}<\frac{1}{3^{n}n!}=b_n\\ b_{n+1}<b_n\end{array}$

(ii) $\begin{array}{l}\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }\frac{1}{3^{n}n!}\\ =\frac{1}{3^{n}n!}\to 0\text{as}n\to \infty \end{array}$

So, by this it can be said that the given series is convergent.

For how many terms itneed the approximation use alternating series estimation

  $\begin{array}{l}s={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{3^{n}n!}}\\ =\frac{-1}{3^{1}1!}+\frac{1}{3^{2}2!}-\frac{1}{3^{3}3!}+\frac{1}{3^{4}4!}-\frac{1}{3^{5}5!}+\mathrm{...............}\end{array}$

Notice that

  $\begin{array}{l}{b}_5=\frac{1}{3^{5}5!}\\ =\frac{1}{29160}<\frac{1}{20000}=0.00005\end{array}$

And

  $\begin{array}{l}{s}_4=\frac{-1}{3^{1}1!}+\frac{1}{3^{2}2!}-\frac{1}{3^{3}3!}+\frac{1}{3^{4}4!}\\ =-\frac{1}{3}+\frac{1}{18}-\frac{1}{162}+\frac{1}{1944}\\ =-0.283436\end{array}$

By the alternating series estimation theorem then

  $\begin{array}{l}\left|s-s_n\right|\le b_{n+1}\\ \left|s-s_n\right|=\text{error}\\ \left|s-s_4\right|\le b_5<0.00005\end{array}$

This error of less than $0.00005$ does not affect the four decimal place,

  $s=-0.2834$ correct to four decimal places.

So, the sum of the series is $s=-0.2834$ correct to four decimal places.

Conclusion:

The sum of the series is $s=-0.2834$ correct to four decimal places.