Chapter 8.4, Problem 19E

To determine

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

Answer to Problem 19E

The sum of the series is $s=0.0676$ correct to four decimal places.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n-1}{n}^2}{{10}^n}}$

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 ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n-1}{n}^2}{{10}^n}}$

Firstly observe that the series is convergent by alternating series test

Here,

  $b_n=\frac{n^2}{{10}^n}$

Then

  $b_{n+1}=\frac{{\left(n+1\right)}^2}{{10}^{n+1}}$

(i) $\begin{array}{l}{b}_{n+1}=\frac{{\left(n+1\right)}^2}{{10}^{n+1}}<\frac{n^2}{{10}^n}=b_n\\ b_{n+1}<b_n\end{array}$

(ii) $\begin{array}{l}\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }\frac{n^2}{{10}^n}\\ =\frac{n^2}{{10}^n}\to 0\text{as}n\to \infty \end{array}$

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

For how many terms itneed in the approximation use alternating series estimation

  $\begin{array}{l}s={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n-1}{n}^2}{{10}^n}}\\ =\frac{1}{10}-\frac{2^2}{{10}^2}+\frac{3^2}{{10}^3}-\frac{4^2}{{10}^4}+\frac{5^2}{{10}^5}-\frac{6^2}{{10}^6}+\mathrm{................}\end{array}$

Notice that

  $\begin{array}{l}{b}_6=\frac{6^2}{{10}^6}\\ =0.000036\end{array}$

And

  $\begin{array}{l}{s}_5=\frac{1}{10}-\frac{2^2}{{10}^2}+\frac{3^2}{{10}^3}-\frac{4^2}{{10}^4}+\frac{5^2}{{10}^5}\\ =0.1-0.04+0.009-0.0016+0.00025\\ =0.06765\end{array}$

By the alternating series estimation theorem then

  $\begin{array}{l}\left|s-s_n\right|\le b_{n+1}\\ \left|s-s_n\right|=\text{error}\\ \left|s-s_5\right|\le b_6<0.000036\end{array}$

This error of less than $0.000036$ does not affect the four decimal place,

  $s=0.0676$ correct to four decimal places.

So, the sum of the series is $s=0.0676$ correct to four decimal places.

Conclusion:

The sum of the series is $s=0.0676$ correct to four decimal places.