So, we know that the series will converge if,

$$6\left| {4x - 1} \right| < 1\hspace{0.25in} \to \hspace{0.25in}24\left| {x - \frac{1}{4}} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x - \frac{1}{4}} \right| < \frac{1}{{24}}$$ Show Step 3

So, from the previous step we see that the radius of convergence is $\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}$.

Show Step 4

Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2.

$$- \frac{1}{{24}} < x - \frac{1}{4} < \frac{1}{{24}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{5}{{24}} < x < \frac{7}{{24}}$$ Show Step 5

To finalize the interval of convergence we need to check the end points of the inequality from Step 4.

$\displaystyle x = \frac{5}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( { - \frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{{6{{\left( { - 1} \right)}^{n - 1}}}}{n}}$

$\displaystyle x = \frac{7}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( {\frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{1}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{6}{n}}$

Now, the first series is an alternating harmonic series which we know converges (or you could just do a quick Alternating Series Test to verify this) and the second series diverges by the $p$-series test.

Show Step 6

The interval of convergence is below and for summary purposes the radius of convergence is also shown.

$$\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{5}{{24}} \le x < \frac{7}{{24}}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}$$