Chapter 8.3, Problem 14E

To determine

To find: whether the series is convergent or divergent.

Answer to Problem 14E

The series is convergent

Explanation of Solution

Given information: The series is $1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\mathrm{...}$

Given, $1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\mathrm{...}$

Let, ${\displaystyle \sum _{n=1}^{\infty }{u}_n=}1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\mathrm{...}+\frac{1}{n\sqrt{n}}$

So,

  $\begin{array}{l}{u}_n=\frac{1}{n\sqrt{n}}\\ \Rightarrow u_n={\left(\frac{1}{n}\right)}^{\frac{3}{2}}\end{array}$

Now we find convergence or divergence of $u_n=\frac{1}{n^{\frac{3}{2}}}$

By the theorem of p series we can say that the series is convergent where $\frac{3}{2}(p\text{ say)}\ge 1$ .

By the definition since $u_n$ is convergent then the partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }{u}_n}$ is also convergent .

Hence , the series is convergent.