91Ó°ÊÓ

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.

The given series is $\underset{1}{\overset{∞}{∑}}\frac{1}{n{\left(1+ \ln n\right)}^{\frac{3}{2}}}$.

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

Now solve the integral is as follows:

Let $t=1+\ln n$,$dt=\frac{1}{n} dn$

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

$\underset{}{\overset{∞}{∫}}\frac{dt}{t^{\frac{3}{2}}}=\underset{}{\overset{∞}{∫}}{t}^{-\frac{3}{2}} dt$

Solve the integral is as follows:

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

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

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