2) Find the interval of convergence if \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n} \cdot 3^{n}}{4^{n}(n+1)} \]

Step 1 ：Rewrite the series as \(\sum_{n=1}^\infty a_n\), where \(a_n = (-1)^n \frac{(x-1)^n \cdot 3^n}{4^n(n+1)}\)

Step 2 ：Apply the Ratio Test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{(x-1)^{n+1} \cdot 3^{n+1}}{4^{n+1}(n+2)}}{(-1)^n \frac{(x-1)^n \cdot 3^n}{4^n(n+1)}} \right|\)

Step 3 ：Simplify the expression: \(\lim_{n \to \infty} \left| \frac{(x-1) \cdot 3}{4} \cdot \frac{n+1}{n+2} \right|\)

Step 4 ：Take the limit as n approaches infinity: \(\left| \frac{3(x-1)}{4} \right|\)

Step 5 ：For the series to converge, the result of the Ratio Test must be less than 1: \(\left| \frac{3(x-1)}{4} \right| < 1\)

Step 6 ：Solve the inequality: \(-1 < \frac{3(x-1)}{4} < 1\)

Step 7 ：Multiply all sides by 4: \(-4 < 3(x-1) < 4\)

Step 8 ：Distribute the 3: \(-4 < 3x-3 < 4\)

Step 9 ：Add 3 to all sides: \(-1 < 3x < 7\)

Step 10 ：Divide all sides by 3: \(-\frac{1}{3} < x < \frac{7}{3}\)

Step 11 ：The interval of convergence is \(\boxed{\left(-\frac{1}{3}, \frac{7}{3}\right)}\)