For [latex]k = 1[/latex]: [latex]T_1 = \frac{1}{1} - \ln(1) = 1 - 0 = 1[/latex]

For [latex]k = 2[/latex]: [latex]T_2 = \left(\frac{1}{1} + \frac{1}{2}\right) - \ln(2) = 1.5 - 0.693147... \approx 0.806853[/latex]

For [latex]k = 3[/latex]: [latex]T_3 = \left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right) - \ln(3) = 1.833333... - 1.098612... \approx 0.734721[/latex]

For [latex]k = 4[/latex]: [latex]T_4 = \left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) - \ln(4) = 2.083333... - 1.386294... \approx 0.697039[/latex]

For [latex]k = 5[/latex]: [latex]T_5 = \left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right) - \ln(5) = 2.283333... - 1.609438... \approx 0.673895[/latex]

For [latex]k = 10[/latex]: [latex]T_{10} = \sum_{n=1}^{10}\frac{1}{n} - \ln(10) = 2.928968... - 2.302585... \approx 0.626383[/latex]

For [latex]k = 100[/latex]: [latex]T_{100} = \sum_{n=1}^{100}\frac{1}{n} - \ln(100) = 5.187378... - 4.605170... \approx 0.582207[/latex]

Notice that [latex]T_k[/latex] appears to be decreasing and approaching a limit around [latex]0.577.[/latex]

1. To show [latex]{T_k}[/latex] is monotone decreasing, we need to show [latex]T_{k+1} < T_k[/latex], or equivalently, [latex]T_k - T_{k+1} > 0[/latex].

   [latex]T_k - T_{k+1} = \left(\sum_{n=1}^{k}\frac{1}{n} - \ln k\right) - \left(\sum_{n=1}^{k+1}\frac{1}{n} - \ln(k+1)\right)[/latex]

   [latex]= \sum_{n=1}^{k}\frac{1}{n} - \ln k - \sum_{n=1}^{k+1}\frac{1}{n} + \ln(k+1)[/latex]

   [latex]= -\frac{1}{k+1} + \ln(k+1) - \ln k[/latex]

   [latex]= \ln\left(\frac{k+1}{k}\right) - \frac{1}{k+1}[/latex]

   [latex]= \ln\left(1 + \frac{1}{k}\right) - \frac{1}{k+1}[/latex]

   Now we need to show that [latex]\ln\left(1 + \frac{1}{k}\right) > \frac{1}{k+1}[/latex].

   Let [latex]f(x) = \ln(1+x) - \frac{x}{1+x}[/latex] for [latex]x > 0[/latex]. We need to show [latex]f\left(\frac{1}{k}\right) > 0[/latex].

   Taking the derivative: [latex]f'(x) = \frac{1}{1+x} - \frac{(1+x) - x}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{1+x-1}{(1+x)^2} = \frac{x}{(1+x)^2}[/latex]

   Since [latex]x > 0[/latex], we have [latex]f'(x) > 0[/latex], so [latex]f(x)[/latex] is increasing for [latex]x > 0[/latex].

   Also, [latex]f(0) = \ln(1) - 0 = 0[/latex].

   Therefore, for [latex]x > 0[/latex], we have [latex]f(x) > f(0) = 0[/latex], which means [latex]\ln(1+x) > \frac{x}{1+x}[/latex].

   Setting [latex]x = \frac{1}{k}[/latex], we get [latex]\ln\left(1 + \frac{1}{k}\right) > \frac{1}{k+1}[/latex].

   Therefore, [latex]T_k - T_{k+1} > 0[/latex], so [latex]{T_k}[/latex] is monotone decreasing.

2. We can express [latex]\ln k = \int_1^k \frac{1}{x} dx[/latex].

   Now, consider the Riemann sum approximation using right endpoints: [latex]\sum_{n=1}^{k-1} \frac{1}{n+1} \cdot 1 = \sum_{n=2}^{k} \frac{1}{n}[/latex]

   This is a right Riemann sum for [latex]\int_1^k \frac{1}{x} dx[/latex] with subintervals [latex][n, n+1][/latex] for [latex]n = 1, 2, ..., k-1[/latex].

   Since [latex]f(x) = \frac{1}{x}[/latex] is decreasing on [latex][1,k][/latex], the right Riemann sum underestimates the integral: [latex]\sum_{n=2}^{k} \frac{1}{n} < \int_1^k \frac{1}{x} dx = \ln k[/latex]

   Therefore: [latex]\sum_{n=1}^{k} \frac{1}{n} = 1 + \sum_{n=2}^{k} \frac{1}{n} < 1 + \ln k[/latex]

   This gives us: [latex]T_k = \sum_{n=1}^{k} \frac{1}{n} - \ln k < 1 + \ln k - \ln k = 1[/latex]

   For a tighter bound, using left Riemann sums: [latex]\sum_{n=1}^{k-1} \frac{1}{n} \cdot 1 = \sum_{n=1}^{k-1} \frac{1}{n}[/latex]

   Since [latex]f(x) = \frac{1}{x}[/latex] is decreasing, the left Riemann sum overestimates the integral: [latex]\sum_{n=1}^{k-1} \frac{1}{n} > \int_1^k \frac{1}{x} dx = \ln k[/latex]

   Therefore: [latex]\sum_{n=1}^{k} \frac{1}{n} = \sum_{n=1}^{k-1} \frac{1}{n} + \frac{1}{k} > \ln k + \frac{1}{k}[/latex]

   This gives us: [latex]T_k = \sum_{n=1}^{k} \frac{1}{n} - \ln k > \ln k + \frac{1}{k} - \ln k = \frac{1}{k} > 0[/latex]

   Therefore, [latex]T_k > 0[/latex] for all [latex]k \geq 1[/latex], so [latex]{T_k}[/latex] is bounded below by zero.

3. From parts (a) and (b), we have shown that:

   • The sequence [latex]{T_k}[/latex] is monotone decreasing
   • The sequence [latex]{T_k}[/latex] is bounded below by zero

   By the Monotone Convergence Theorem, since [latex]{T_k}[/latex] is monotone decreasing and bounded below, the sequence [latex]{T_k}[/latex] converges to some limit.

   We define this limit as Euler’s constant: [latex]\gamma = \lim_{k \to \infty} T_k = \lim_{k \to \infty} \left(\sum_{n=1}^{k}\frac{1}{n} - \ln k\right)[/latex]

   Therefore, [latex]\gamma \approx 0.5772156649...[/latex] (Euler’s constant).

1. We need to show [latex]\ln(k+1) - \ln k < \frac{1}{k}[/latex], which is equivalent to [latex]\ln\left(\frac{k+1}{k}\right) < \frac{1}{k}[/latex], or [latex]\ln\left(1 + \frac{1}{k}\right) < \frac{1}{k}[/latex].

   Consider the function [latex]g(x) = \ln(1+x) - x[/latex] for [latex]x > 0[/latex].

   Taking the derivative: [latex]g'(x) = \frac{1}{1+x} - 1 = \frac{1-(1+x)}{1+x} = \frac{-x}{1+x}[/latex]

   Since [latex]x > 0[/latex], we have [latex]g'(x) < 0[/latex], so [latex]g(x)[/latex] is decreasing for [latex]x > 0[/latex].

   Also, [latex]g(0) = \ln(1) - 0 = 0[/latex].

   Therefore, for [latex]x > 0[/latex], we have [latex]g(x) < g(0) = 0[/latex], which means [latex]\ln(1+x) < x[/latex].

   Setting [latex]x = \frac{1}{k}[/latex], we get [latex]\ln\left(1 + \frac{1}{k}\right) < \frac{1}{k}[/latex].

   Therefore, [latex]\ln(k+1) - \ln k < \frac{1}{k}[/latex].

2. From the solution to Problem 2(a), we found: [latex]T_k - T_{k+1} = \ln(k+1) - \ln k - \frac{1}{k+1}[/latex]

   From part (a), we know [latex]\ln(k+1) - \ln k < \frac{1}{k}[/latex].

   Therefore: [latex]T_k - T_{k+1} = \ln(k+1) - \ln k - \frac{1}{k+1} < \frac{1}{k} - \frac{1}{k+1}[/latex]

3. We can write: [latex]T_k - T_j = (T_k - T_{k+1}) + (T_{k+1} - T_{k+2}) + (T_{k+2} - T_{k+3}) + \cdots + (T_{j-1} - T_j)[/latex]

   This is a telescoping sum. From part (b), each term satisfies:

   • [latex]T_k - T_{k+1} < \frac{1}{k} - \frac{1}{k+1}[/latex]
   • [latex]T_{k+1} - T_{k+2} < \frac{1}{k+1} - \frac{1}{k+2}[/latex]
   • [latex]T_{k+2} - T_{k+3} < \frac{1}{k+2} - \frac{1}{k+3}[/latex]
   • [latex]\dots[/latex]
   • [latex]T_{j-1} - T_j < \frac{1}{j-1} - \frac{1}{j}[/latex]

   Adding all these inequalities: [latex]T_k - T_j < \left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{k+1} - \frac{1}{k+2}\right) + \cdots + \left(\frac{1}{j-1} - \frac{1}{j}\right)[/latex]

   The right side is a telescoping sum that simplifies to: [latex]T_k - T_j < \frac{1}{k} - \frac{1}{j}[/latex]

4. From part (c), we have [latex]T_k - T_j < \frac{1}{k} - \frac{1}{j}[/latex] for any [latex]j > k[/latex].

   Taking the limit as [latex]j \to \infty[/latex]: [latex]\lim_{j \to \infty} (T_k - T_j) \leq \lim_{j \to \infty} \left(\frac{1}{k} - \frac{1}{j}\right)[/latex]

   Since [latex]\lim_{j \to \infty} T_j = \gamma[/latex] and [latex]\lim_{j \to \infty} \frac{1}{j} = 0[/latex]: [latex]T_k - \gamma \leq \frac{1}{k} - 0 = \frac{1}{k}[/latex]

   We already showed in Problem 2(b) that [latex]T_k > 0[/latex], and since [latex]{T_k}[/latex] is decreasing and converges to [latex]\gamma[/latex], we have [latex]T_k > \gamma[/latex].

   Therefore: [latex]0 < T_k - \gamma \leq \frac{1}{k}[/latex]

5. From part (d), we know that [latex]|T_k - \gamma| = T_k - \gamma \leq \frac{1}{k}[/latex].

   To achieve accuracy within [latex]0.001[/latex], we need [latex]\frac{1}{k} \leq 0.001[/latex], which means [latex]k \geq 1000[/latex].

   Let’s calculate [latex]T_{1000}[/latex]: [latex]T_{1000} = \sum_{n=1}^{1000}\frac{1}{n} - \ln(1000)[/latex]

   Using numerical computation: [latex]\sum_{n=1}^{1000}\frac{1}{n} \approx 7.485470861[/latex] [latex]\ln(1000) \approx 6.907755279[/latex]

   Therefore: [latex]T_{1000} \approx 7.485470861 - 6.907755279 \approx 0.577715582[/latex]

   Since [latex]T_{1000} - \gamma \leq \frac{1}{1000} = 0.001[/latex], we have: [latex]\gamma \geq T_{1000} - 0.001 \approx 0.577715582 - 0.001 = 0.576715582[/latex] [latex]\gamma \leq T_{1000} \approx 0.577715582[/latex]

   Therefore, to an accuracy of within [latex]0.001[/latex]: [latex]\gamma \approx 0.5777[/latex]

   The actual value of Euler’s constant is [latex]\gamma \approx 0.5772156649...[/latex], which confirms our estimate is accurate to within the required tolerance.