The Euler–Mascheroni Constant

$\gamma_{n+1} - \gamma_n = (H_{n+1} - \ln(n+1)) - (H_n - \ln n) = \frac{1}{n+1} - (\ln(n+1) - \ln n)$. By the left half of the hint inequality, $\ln(n+1) - \ln n > \frac{1}{n+1}$, so this difference is negative: $\gamma_n$ is strictly decreasing.

The lower bound appears by splitting $\ln n$: $\gamma_n = H_n - \sum_{k=1}^{n-1}(\ln(k+1) - \ln k) = \frac{1}{n} + \sum_{k=1}^{n-1}\left(\frac{1}{k} - \ln\left(1+\frac{1}{k}\right)\right)$, and each bracket is positive by the right half of the hint inequality, $\ln\left(1+\frac{1}{k}\right) < \frac{1}{k}$. Hence $\gamma_n > \frac{1}{n} > 0$.

A decreasing sequence bounded below converges by the monotone convergence theorem. Its limit is the Euler–Mascheroni constant $\gamma := \lim_{n\to\infty} \gamma_n$, with $\gamma \ge 0$ (in fact $\gamma \approx 0.5772$).

Numerical check ($n = 50$): $\gamma_{50} = 4.499205338 - 3.912023005 = 0.5871823329$, and $\gamma_{51} = \gamma_{50} - (\ln(51/50) - 1/51) \approx 0.5871823329 - 0.0001948 = 0.5869875$ — matching the caption's drop of $0.0001948$.

Expanding the partial sum, $\sum_{k=1}^{n} \ln\left(1+\frac{1}{k}\right) = \sum_{k=1}^{n}(\ln(k+1) - \ln k) = \ln(n+1)$ telescopes, so $\sum_{k=1}^{n}\left(\frac{1}{k} - \ln\left(1+\frac{1}{k}\right)\right) = H_n - \ln(n+1)$.

Since $H_n - \ln(n+1) = \gamma_n - \ln\left(1+\frac{1}{n}\right)$, with $\gamma_n \to \gamma$ (problem 1) and $\ln\left(1+\frac{1}{n}\right) \to 0$, the partial sums converge to $\gamma$. Hence the series sums to $\gamma$.

Each term $d_k := \frac{1}{k} - \ln\left(1+\frac{1}{k}\right)$ is positive by $\ln\left(1+\frac{1}{k}\right) < \frac{1}{k}$, so this says a series of positive terms sums to $\gamma$ — this is exactly why the gap between the diverging harmonic series and $\ln n$ converges: the total of these differences is finite.

Numerical check ($n = 100$): $H_{100} - \ln 101 \approx 5.187377518 - 4.615120517 = 0.5722570008$ — matching the caption and approaching $\gamma \approx 0.5772$ (the gap $\approx 0.005$ is the tail $\sum_{k>100} d_k$).

By problem 2, $\gamma = \sum_{k=1}^{\infty} d_k$ with every $d_k > 0$.

The first 10 terms telescope to $\sum_{k=1}^{10} d_k = H_{10} - \ln 11 \approx 2.9289682539 - 2.3978952728 = 0.5310729812$.

Since $d_k > 0$, this is already a lower bound: $\gamma > 0.5310729812 > 0.5$.

To bound the tail from above, estimate $d_k$ from above.

Write $\ln\left(1+\frac{1}{k}\right) = \int_k^{k+1} \frac{dx}{x}$; by convexity of $\frac{1}{x}$, $f(k+t) + f(k+1-t) \ge 2f(k+\tfrac12)$ for $0 \le t \le 1$, so integrating gives $\int_k^{k+1} \frac{dx}{x} \ge \frac{1}{k+1/2} = \frac{2}{2k+1}$.

Hence $d_k \le \frac{1}{k} - \frac{2}{2k+1} = \frac{1}{k(2k+1)} < \frac{1}{2k^2}$.

Now $\sum_{k=11}^{\infty} d_k < \frac{1}{2}\sum_{k=11}^{\infty} \frac{1}{k^2} < \frac{1}{2}\sum_{k=11}^{\infty} \frac{1}{k(k-1)} = \frac{1}{2}\sum_{k=11}^{\infty}\left(\frac{1}{k-1} - \frac{1}{k}\right) = \frac{1}{2}\cdot\frac{1}{10} = 0.05$ (using $\frac{1}{k^2} < \frac{1}{k(k-1)}$ for $k \ge 2$, and the last sum telescopes).

Therefore $0.5310729812 < \gamma < 0.5310729812 + 0.05 = 0.5810729812$, so certainly $0.5 < \gamma < 0.6$. The true value $\gamma \approx 0.5772$ lies inside this interval.