Chapter 8.4, Problem 16E

To determine

To calculate:The number of terms of the series needed to add in order to get the sum to indicated accuracy?

Answer to Problem 16E

The given series is convergent and the six terms we needed to add in order to get the sum to the indicated $0.01$ accuracy.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}n}{e}^{-n}\left(\left|\text{error}\right|<0.01\right)$

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

For series convergent use alternating series test

Here,

  $b_n=n{e}^{-n}$

Then

  $b_{n+1}=\left(n+1\right)e^{-\left(n+1\right)}$

(i) $\begin{array}{l}{b}_{n+1}=\left(n+1\right)e^{-\left(n+1\right)}<n{e}^{-n}=b_n\\ \text{so}{b}_{n+1}<b_n\end{array}$

(ii) $\begin{array}{l}\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }n{e}^{-n}\\ =n{e}^{-n}\to 0\text{as}n\to \infty \end{array}$

So, by this we can say that the given series is convergent.

For sum to the indicated accuracy use alternating series estimation

  $\begin{array}{l}s={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}n}{e}^{-n}\\ =1\cdot e^{-1}-2\cdot e^{-2}+3\cdot e^{-3}-4\cdot e^{-4}+5\cdot e^{-5}-6\cdot e^{-6}+7\cdot e^{-7}+\mathrm{................}\end{array}$

Notice that

  $\begin{array}{l}{b}_7=7\cdot e^{-7}\\ =0.00638\end{array}$

And

  $\begin{array}{l}{s}_6=1\cdot e^{-1}-2\cdot e^{-2}+3\cdot e^{-3}-4\cdot e^{-4}+5\cdot e^{-5}-6\cdot e^{-6}\\ =\frac{1}{e}-\frac{2}{e^2}+\frac{3}{e^3}-\frac{4}{e^4}+\frac{5}{e^5}-\frac{6}{e^6}\\ =0.3678-0.2706+0.1493-0.073+0.033-0.014\\ =0.1925\end{array}$

By the alternating series estimation theorem then

  $\begin{array}{l}\left|s-s_n\right|\le b_{n+1}\\ \left|s-s_n\right|=\text{error}\\ \left|s-s_6\right|\le b_7<0.01\end{array}$

This error of less than $0.01$ does not affect the first decimal place, so we have

  $s=0.1$ correct to one decimal places.

So, the six terms we needed to add in order to find the sum to the indicated $0.01$ accuracy.

Conclusion:

The given series is convergent and the six terms we needed to add in order to get the sum to the indicated $0.01$ accuracy.