Chapter 10.5, Problem 29E

To determine

Whether the series converges conditionally, converges absolutely, or diverges.

Answer to Problem 29E

The series is conditionally convergent.

Explanation of Solution

Given information:

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

The series is alternating.

Then

Check for absolute convergence by considering the series of absolute values:

  ${\displaystyle {\sum }_1^{\infty }\left|\frac{\pm 1}{1+\sqrt{n}}\right|}={\displaystyle {\sum }_1^{\infty }\frac{1}{1+\sqrt{n}}}$

Such that

  ${\displaystyle {\sum }_1^{\infty }\frac{1}{1+\sqrt{n}}}>{\displaystyle {\sum }_1^{\infty }\frac{1}{\sqrt{n}+\sqrt{n}}}$

Thus,

  $\frac{1}{2\sqrt{n}}$ is divergent.

And

  ${\displaystyle {\sum }_1^{\infty }\frac{1}{1+\sqrt{n}}}$ is also divergent.

Hence,

The series is not absolutely convergent.

Now,

Check for the conditional convergence:

  ${\displaystyle {\sum }_1^{\infty }{\left(-1\right)}^{n+1}{a}_n}={\displaystyle {\sum }_{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{1+\sqrt{n}}}$

Then

  $a_n=\frac{1}{1+\sqrt{n}}$ is decreasing for $n>1$

Such that

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

Therefore,

The series is conditionally convergent.