Answer
The interval of convergence is $[1,2)$.
Work Step by Step
Let ${a_n} = \dfrac{{{{\left( {2x - 3} \right)}^n}}}{{n\ln n}}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{{\left( {2x - 3} \right)}^{n + 1}}}}{{\left( {n + 1} \right)\ln \left( {n + 1} \right)}}\cdot\dfrac{{n\ln n}}{{{{\left( {2x - 3} \right)}^n}}}} \right|$
$ = \left| {2x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{n}{{\left( {n + 1} \right)}}\cdot\dfrac{{\ln n}}{{\ln \left( {n + 1} \right)}}} \right|$
$ = \left| {2x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{\left( {1 + \dfrac{1}{n}} \right)}}} \right|\cdot\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\ln n}}{{\ln \left( {n + 1} \right)}}} \right|$
Using L'Hôpital's Rule:
$\mathop = \limits^{{\rm{L'Hô pital's}}} \left| {2x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{\left( {1 + \dfrac{1}{n}} \right)}}} \right|\cdot\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\dfrac{1}{n}}}{{\dfrac{1}{{n + 1}}}}} \right|$
$ = \left| {2x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{\left( {1 + \dfrac{1}{n}} \right)}}} \right|\cdot\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n + 1}}{n}} \right|$
$ = \left| {2x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{\left( {1 + \dfrac{1}{n}} \right)}}} \right|\cdot\mathop {\lim }\limits_{n \to \infty } \left| {1 + \dfrac{1}{n}} \right| = \left| {2x - 3} \right|$
The series converges if $\rho = \left| {2x - 3} \right| \lt 1$. That is,
$ - 1 \lt 2x - 3 \lt 1$
$2 \lt 2x \lt 4$
$1 \lt x \lt 2$
Next, we check the endpoints: $x=1$ and $x=2$.
1. $x=1$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( {2x - 3} \right)}^n}}}{{n\ln n}} = \mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{n\ln n}}$
Since the sequence $\left\{ {\dfrac{1}{{n\ln n}}} \right\}$ is positive, decreasing and converge to $0$, by the Alternating Series Test, this series converges.
2. $x=2$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( {2x - 3} \right)}^n}}}{{n\ln n}} = \mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{n\ln n}}$
Let $f\left( x \right) = \dfrac{1}{{x\ln x}}$.
For $x \ge 2$, $f$ is a positive, decreasing, and continuous function of $x$, so we can use the Integral Test.
Evaluate the integral: $\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x$.
Let $t = \ln x$. So, $dt = \dfrac{1}{x}dx$. The integral above becomes
$\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x = \mathop \smallint \limits_{\ln 2}^\infty \dfrac{1}{t}{\rm{d}}t = \ln t|_{\ln 2}^\infty = \infty $
Since $\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x$ diverges, by the Integral Test: $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{n\ln n}}$ also diverges.
Thus, the interval of convergence is $[1,2)$.
