1. Find the interval of convergence of the power series \( \sum_{n=1}^{+\infty} \frac{(x-1)^{n}}{n 3^{n}} \)

To find the interval of convergence of the power series \[ \sum_{n=1}^{+\infty} \frac{(x-1)^{n}}{n 3^{n}}, \] we can use the Ratio Test. The general term of the series is \[ a_n = \frac{(x-1)^{n}}{n 3^{n}}. \] We will compute the limit \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \] First, we find \( a_{n+1} \): \[ a_{n+1} = \frac{(x-1)^{n+1}}{(n+1) 3^{n+1}}. \] Now, we compute the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{(x-1)^{n+1}}{(n+1) 3^{n+1}} \cdot \frac{n 3^{n}}{(x-1)^{n}} = \frac{(x-1) \cdot n}{(n+1) \cdot 3}. \] Taking the absolute value, we have: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-1) \cdot n}{(n+1) \cdot 3} \right| = \frac{|x-1| \cdot n}{3(n+1)}. \] Now, we take the limit as \( n \to \infty \): \[ L = \lim_{n \to \infty} \frac{|x-1| \cdot n}{3(n+1)} = \lim_{n \to \infty} \frac{|x-1| \cdot n}{3n + 3} = \lim_{n \to \infty} \frac{|x-1|}{3 + \frac{3}{n}} = \frac{|x-1|}{3}. \] According to the Ratio Test, the series converges if \( L < 1 \): \[ \frac{|x-1|}{3} < 1. \] Multiplying both sides by 3 gives: \[ |x-1| < 3. \] This inequality can be rewritten as: \[ -3 < x - 1 < 3. \] Adding 1 to all parts of the inequality results in: \[ -2 < x < 4. \] Thus, the interval of convergence is \[ (-2, 4). \] Next, we need to check the endpoints \( x = -2 \) and \( x = 4 \). 1. **At \( x = -2 \)**: \[ \sum_{n=1}^{+\infty} \frac{(-2-1)^{n}}{n 3^{n}} = \sum_{n=1}^{+\infty} \frac{(-3)^{n}}{n 3^{n}} = \sum_{n=1}^{+\infty} \frac{(-1)^{n}}{n} = -\sum_{n=1}^{+\infty} \frac{1}{n}. \] This series diverges (it is the harmonic series). 2. **At \( x = 4 \)**: \[ \sum_{n=1}^{+\infty} \frac{(4-1)^{n}}{n 3^{n}} = \sum_{n=1}^{+\infty} \frac{3^{n}}{n 3^{n}} = \sum_{n=1}^{+\infty} \frac{1}{n}. \] This series also diverges. Since both endpoints do not converge, the interval of convergence is \[ \boxed{(-2, 4)}. \]