Question

Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.

$\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{n}{\left(1+ lnn\right)}^{\frac{\mathbf{3}}{\mathbf{2}}}}$

Short answer

The series$\sum _1^{\infty }\frac{1}{n{\left(1+ \ln n\right)}^{\frac{3}{2}}}$ is divergent.

Step-by-step solution

Definition of convergent and divergent.

If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series tend to a limit S, the series is called convergent. If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series don't approach a limit, the series is called divergent.

The limiting value S is called the sum of the series.

Integral test.

The given series is $\sum _1^{\infty }\frac{1}{n{\left(1+ \ln n\right)}^{\frac{3}{2}}}$.

Use the integral in the given series, $\underset{}{\overset{\infty }{\int }}\frac{1}{n{\left(1+\ln n\right)}^{\frac{3}{2}}} dn$.

Now solve the integral is as follows:

Let $t=1+\ln n$,$dt=\frac{1}{n} dn$

Solve integral.

Substitute the values of t and dt into the integral and change the variable from n into t.

$\underset{}{\overset{\infty }{\int }}\frac{dt}{t^{\frac{3}{2}}}=\underset{}{\overset{\infty }{\int }}{t}^{-\frac{3}{2}} dt$

Solve the integral is as follows:

Use the integral formula, $\int x^a dx=\frac{x^{a+1}}{a+1} dx+c$, where c is a constant.

$\begin{array}{rcl}\underset{}{\overset{\infty }{\int }}{t}^{-\frac{3}{2}} dt& =& {\left[\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\right]}^{\infty }\\ & =& {\left[\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}\right]}^{\infty }\\ & =& -2{\left[t^{-\frac{1}{2}}\right]}^{\infty }\\ & =& -2\left[\infty \right]\\ & =& \infty \end{array}$

Hence, the series approaches to infinite therefore the given series diverges.