Chapter 8, Problem 15RE

To determine

To find: Whether the series is convergent or divergent.

Answer to Problem 15RE

The series is convergent.

Explanation of Solution

Given:

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

Calculation:

Since this is an alternating series so use test for alternating series.

Here $b_n=\frac{\sqrt{n}}{n+1}$ .

Let $f\left(x\right)=\frac{\sqrt{x}}{x+1}$ .

Find the derivative of the function $f\left(x\right)=\frac{\sqrt{x}}{x+1}$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{\left(x+1\right)\left(\frac{1}{2}\sqrt{x}\right)-\sqrt{x}}{{\left(x+1\right)}^2}\\ =\frac{\frac{1}{2\sqrt{x}}\left(x+1-2x\right)}{{\left(x+1\right)}^2}\\ =\frac{1-x}{2\sqrt{x}{\left(x+1\right)}^2}\end{array}$

The expression for $x>1$ .

So $\begin{array}{c}f\left(n\right)>f\left(n+1\right)\\ b_{n+1}\le b_n\end{array}$

Simplify $\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{n+1}$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{n+1}\\ =\underset{n\to \infty }{\lim }\frac{\frac{1}{\sqrt{n}}}{1+\frac{1}{n}}\\ =0\end{array}$

Therefore, the series is convergent.