Chapter 10.5, Problem 24E

To determine

Whether the series converges conditionally, converges absolutely, or diverges and bound for truncation error after 99 terms.

Answer to Problem 24E

The series is conditionally convergent.

Truncation error after 99 terms is $\pm 0.0022$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=2}^{\infty }{\left(-1\right)}^{n+1}{\left(0.1\right)}^n}$

The series is alternating.

Thus,

We need to check for absolute convergence by considering the series of absolute values:

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

Now,

Use integral test to determine whether the series converges or not.

Let

  $f\left(x\right)=\frac{1}{x\ln x}$

The function $f\left(x\right)$ is decreasing for $x\ge 2$ .

Now,

Determine whether the following integral converges or not:

  ${\displaystyle {\int }_2^{\infty }\frac{dx}{x\ln x}}={\ln \left(\ln x\right)|}_2^{\infty }=\ln \left(\ln \infty \right)-\ln \left(\ln 2\right)=\infty$

Since the integral diverges, the series of absolute values diverges by the Integral Test and the series ${\displaystyle {\sum }_{n=2}^{\infty }\frac{{\left(-1\right)}^n}{n\ln n}}$ is not absolutely convergent.

Now,

Use Leibniz Test to check the conditional convergence.

Let

  $a_n=\frac{1}{n\ln n}$

Then

  $a_n$ is decreasing for $n\ge 2$

And

  $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{n\ln n}=0$

Therefore,

The series is conditionally convergent.

Then

Truncation error after 99 terms:

  $\pm a_{100}=\frac{\pm 1}{100\ln 100}=\pm 0.0022$