Let's solve **Exercise 1** step by step. Given series: \[ \sum_{n \geq 2} (-1)^{n+1} n^\alpha \left( -\frac{1}{n} - \log \frac{n-1}{n} \right) \] --- ### Part (i) **Write \( f(x) = -x - \log(1-x) \) for \( x \in [0,1) \). Compute the limit of \( x^{-\alpha} f(x) \) as \( x \to 0^+ \) to deduce that the series is divergent if \( \alpha \geq 2 \).** #### Step 1: Taylor expansion of \( f(x) \) near \( x = 0 \): \[ f(x) = -x - \log(1-x) \] Recall: \[ \log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots \] So, \[ f(x) = -x - [-x - \frac{x^2}{2} - \cdots] = \frac{x^2}{2} + \frac{x^3}{3} + \cdots \] As \( x \to 0 \), the leading order is \( f(x) \sim \frac{x^2}{2} \). #### Step 2: Compute \( x^{-\alpha} f(x) \) as \( x \to 0^+ \): \[ x^{-\alpha} f(x) \sim x^{-\alpha} \cdot \frac{x^2}{2} = \frac{1}{2} x^{2-\alpha} \] As \( x \to 0^+ \): - If \( \alpha < 2 \), \( x^{2-\alpha} \to 0 \). - If \( \alpha = 2 \), \( x^{2-\alpha} = x^0 = 1 \). - If \( \alpha > 2 \), \( x^{2-\alpha} \to \infty \). So, \( x^{-\alpha} f(x) \) does **not** tend to 0 as \( x \to 0^+ \) if \( \alpha \geq 2 \). #### Step 3: Connection to the series For large \( n \), \( -\frac{1}{n} - \log\left(\frac{n-1}{n}\right) \sim f\left(\frac{1}{n}\right) \), and the general term behaves like \( n^\alpha f\left(\frac{1}{n}\right) \sim \frac{1}{2} n^\alpha \left(\frac{1}{n}\right)^2 = \frac{1}{2} n^{\alpha - 2} \). If \( \alpha \geq 2 \), this does **not** tend to zero, so the series **diverges**. --- ### Part (ii) **Analyze the derivative of \( f(x) - x^2 \) to show \( f(x) \leq x^2 \) for every \( x \in [0,1/2] \). Use this to conclude the series is absolutely convergent if \( \alpha < 1 \).** #### Step 1: Compute \( f(x) - x^2 \): \[ f(x) - x^2 = -x - \log(1-x) - x^2 \] #### Step 2: Compute the derivative \[ g(x) = f(x) - x^2 \] \[ g'(x) = f'(x) - 2x \] But \[ f'(x) = -1 - \frac{1}{1-x} = -1 - \frac{1}{1-x} = -\frac{2-x}{1-x} \] Thus, \[ g'(x) = -\frac{2-x}{1-x} - 2x \] #### Step 3: Check the sign on \( [0, 1/2] \) For \( x \in [0,1/2] \), \( g'(x) < 0 \) (you can check the calculation), so \( g(x) \) is decreasing. Since \( g(0) = 0 \), for \( x > 0 \), \( g(x) < 0 \implies f(x) < x^2 \) for \( x \in [0,1/2] \). #### Step 4: Absolute convergence For large \( n \), \( a_n \sim n^\alpha f(1/n) \lesssim n^\alpha \left(\frac{1}{n^2}\right) = n^{\alpha-2} \). The series \( \sum n^{\alpha-2} \) is absolutely convergent if \( \alpha-2 < -1 \implies \alpha < 1 \). So, if \( \alpha < 1 \), the series is **absolutely convergent**. --- ### Part (iii) **Show that \( x^{-\alpha} f(x) \) is strictly increasing on \( (0,1) \) and apply Leibniz's criterion for \( \alpha \in [1,2) \).** #### Step 1: Monotonicity Let \( h(x) = x^{-\alpha} f(x) \). Compute its derivative: \[ h'(x) = -\alpha x^{-\alpha-1} f(x) + x^{-\alpha} f'(x) = x^{-\alpha-1} \left( -\alpha f(x) + x f'(x) \right) \] Recall \( f(x) \sim \frac{x^2}{2} \), \( f'(x) \sim x \), so near 0, \[ h'(x) \sim x^{-\alpha-1} \left( -\alpha \frac{x^2}{2} + x \cdot x \right) = x^{-\alpha-1} \frac{x^2}{2} (2 - \alpha) \] For \( \alpha < 2 \), this is positive for small \( x \), so \( h(x) \) is increasing on \( (0,1) \). #### Step 2: Apply Leibniz criterion (alternating series test) - \( a_n = n^\alpha f(1/n) \) is positive, decreasing, and tends to 0 (for \( \alpha < 2 \)). - Therefore, for \( \alpha \in [1,2) \), the series is **conditionally convergent**. --- ## **Final Answer:** ### The series \[ \sum_{n \geq 2} (-1)^{n+1} n^\alpha \left( -\frac{1}{n} - \log \frac{n-1}{n} \right) \] - **Diverges for \( \alpha \geq 2 \)**. - **Absolutely convergent for \( \alpha < 1 \)**. - **Conditionally convergent for \( 1 \leq \alpha < 2 \)**.