1. **Problem statement:** We want to determine for which values of $x$ the series $$\sum_{n=1}^{\infty} \frac{1}{n} \left(1+\frac{1}{x}\right)^n$$ is absolutely convergent, conditionally convergent, or divergent. 2. **Recall the series:** This is a power series in terms of $$a = 1 + \frac{1}{x}$$ with general term $$\frac{1}{n} a^n$$. 3. **Important rule:** The series $$\sum_{n=1}^\infty \frac{a^n}{n}$$ converges if and only if $$|a| \leq 1$$, and it converges absolutely if $$|a| < 1$$. If $$a = -1$$, the series converges conditionally (alternating harmonic series). If $$|a| > 1$$, the series diverges. 4. **Analyze $$a = 1 + \frac{1}{x}$$:** - Absolute convergence requires $$|1 + \frac{1}{x}| < 1$$. - Conditional convergence occurs if $$1 + \frac{1}{x} = -1$$. - Divergence if $$|1 + \frac{1}{x}| > 1$$ and not equal to -1. 5. **Solve absolute convergence inequality:** $$|1 + \frac{1}{x}| < 1$$ This means: $$-1 < 1 + \frac{1}{x} < 1$$ Subtract 1 throughout: $$-2 < \frac{1}{x} < 0$$ 6. **Solve inequalities for $$x$$:** - From $$\frac{1}{x} < 0$$, we get $$x < 0$$. - From $$\frac{1}{x} > -2$$: If $$x < 0$$, then multiplying by $$x$$ (negative) reverses inequality: $$1 > -2x$$ Divide by -2 (negative), reverse inequality: $$x > -\frac{1}{2}$$ 7. **Combine:** $$-\frac{1}{2} < x < 0$$ for absolute convergence. 8. **Check conditional convergence:** Set $$1 + \frac{1}{x} = -1$$ $$\frac{1}{x} = -2$$ $$x = -\frac{1}{2}$$ At $$x = -\frac{1}{2}$$, the series is $$\sum \frac{(-1)^n}{n}$$ which converges conditionally. 9. **Check divergence:** For $$x > 0$$ or $$x \leq -\frac{1}{2}$$ (except $$x = -\frac{1}{2}$$), the series diverges. **Final answer:** - Absolutely convergent for $$-\frac{1}{2} < x < 0$$ - Conditionally convergent at $$x = -\frac{1}{2}$$ - Divergent otherwise.