The Euler–Mascheroni Constant

The harmonic numbers HnH_n diverge like ln⁡n\ln n, but their gap γn=Hn−ln⁡n\gamma_n = H_n - \ln n settles at γ≈0.5772\gamma \approx 0.5772 — the secret is the convergent series ∑(1n−ln⁡(1+1n))\sum (\frac1n - \ln(1+\frac1n)).

γ=lim⁡n→∞(Hn−ln⁡n)≈0.5772\gamma = \lim_{n\to\infty}(H_n - \ln n) \approx 0.5772
γ=∑n≥1(1n−ln⁡ ⁣(1+1n))\gamma = \sum_{n\ge1}\left(\frac1n - \ln\!\left(1+\frac1n\right)\right)

Part of these learning paths

Examples & Exercises

Show that γn=Hn−ln⁡n\gamma_n = H_n - \ln n is decreasing and bounded below, hence convergent. Hint: use 1n+1<ln⁡(n+1)−ln⁡n<1n\frac{1}{n+1} < \ln(n+1) - \ln n < \frac{1}{n}.

γn+1−γn=1n+1−(ln⁡(n+1)−ln⁡n)<0\gamma_{n+1} - \gamma_n = \frac{1}{n+1} - \bigl(\ln(n+1) - \ln n\bigr) < 0
n=50n = 50: H50≈4.499205338H_{50} \approx 4.499205338, ln⁡50≈3.912023005\ln 50 \approx 3.912023005, γ50≈0.5871823329\gamma_{50} \approx 0.5871823329 — one step down drops by γ50−γ51=ln⁡(51/50)−151≈0.0001948>0\gamma_{50} - \gamma_{51} = \ln(51/50) - \frac{1}{51} \approx 0.0001948 > 0.
Show proof
  1. γn+1−γn=(Hn+1−ln⁡(n+1))−(Hn−ln⁡n)=1n+1−(ln⁡(n+1)−ln⁡n)\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>1n+1\ln(n+1) - \ln n > \frac{1}{n+1}, so this difference is negative: γn\gamma_n is strictly decreasing.
  2. The lower bound appears by splitting ln⁡n\ln n: γn=Hn−∑k=1n−1(ln⁡(k+1)−ln⁡k)=1n+∑k=1n−1(1k−ln⁡(1+1k))\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⁡(1+1k)<1k\ln\left(1+\frac{1}{k}\right) < \frac{1}{k}. Hence γn>1n>0\gamma_n > \frac{1}{n} > 0.
  3. A decreasing sequence bounded below converges by the monotone convergence theorem. Its limit is the Euler–Mascheroni constant γ:=lim⁡n→∞γn\gamma := \lim_{n\to\infty} \gamma_n, with γ≥0\gamma \ge 0 (in fact γ≈0.5772\gamma \approx 0.5772).
  4. Numerical check (n=50n = 50): γ50=4.499205338−3.912023005=0.5871823329\gamma_{50} = 4.499205338 - 3.912023005 = 0.5871823329, and γ51=γ50−(ln⁡(51/50)−1/51)≈0.5871823329−0.0001948=0.5869875\gamma_{51} = \gamma_{50} - (\ln(51/50) - 1/51) \approx 0.5871823329 - 0.0001948 = 0.5869875 — matching the caption's drop of 0.00019480.0001948.

Show that ∑n=1∞(1n−ln⁡(1+1n))=γ\sum_{n=1}^{\infty}\left(\frac{1}{n} - \ln\left(1+\frac{1}{n}\right)\right) = \gamma (a telescoping sum).

∑k=1n(1k−ln⁡(1+1k))=Hn−ln⁡(n+1)=γn−ln⁡(1+1n)\sum_{k=1}^{n}\left(\frac{1}{k} - \ln\left(1+\frac{1}{k}\right)\right) = H_n - \ln(n+1) = \gamma_n - \ln\left(1+\frac{1}{n}\right)
n=100n = 100: partial sum =H100−ln⁡101≈5.187377518−4.615120517=0.5722570008= H_{100} - \ln 101 \approx 5.187377518 - 4.615120517 = 0.5722570008 — converging toward γ≈0.5772\gamma \approx 0.5772.
Show proof
  1. Expanding the partial sum, ∑k=1nln⁡(1+1k)=∑k=1n(ln⁡(k+1)−ln⁡k)=ln⁡(n+1)\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 ∑k=1n(1k−ln⁡(1+1k))=Hn−ln⁡(n+1)\sum_{k=1}^{n}\left(\frac{1}{k} - \ln\left(1+\frac{1}{k}\right)\right) = H_n - \ln(n+1).
  2. Since Hn−ln⁡(n+1)=γn−ln⁡(1+1n)H_n - \ln(n+1) = \gamma_n - \ln\left(1+\frac{1}{n}\right), with γn→γ\gamma_n \to \gamma (problem 1) and ln⁡(1+1n)→0\ln\left(1+\frac{1}{n}\right) \to 0, the partial sums converge to γ\gamma. Hence the series sums to γ\gamma.
  3. Each term dk:=1k−ln⁡(1+1k)d_k := \frac{1}{k} - \ln\left(1+\frac{1}{k}\right) is positive by ln⁡(1+1k)<1k\ln\left(1+\frac{1}{k}\right) < \frac{1}{k}, so this says a series of positive terms sums to γ\gamma — this is exactly why the gap between the diverging harmonic series and ln⁡n\ln n converges: the total of these differences is finite.
  4. Numerical check (n=100n = 100): H100−ln⁡101≈5.187377518−4.615120517=0.5722570008H_{100} - \ln 101 \approx 5.187377518 - 4.615120517 = 0.5722570008 — matching the caption and approaching γ≈0.5772\gamma \approx 0.5772 (the gap ≈0.005\approx 0.005 is the tail ∑k>100dk\sum_{k>100} d_k).

Prove that 0.5<γ<0.60.5 < \gamma < 0.6. Hint: compute the first 10 terms of γ=∑k=1∞dk\gamma = \sum_{k=1}^{\infty} d_k (dk=1k−ln⁡(1+1k)d_k = \frac{1}{k} - \ln\left(1+\frac{1}{k}\right)) directly and bound the tail rigorously with integral estimates.

γ=∑k=1∞dk,dk:=1k−ln⁡(1+1k)>0\gamma = \sum_{k=1}^{\infty} d_k, \qquad d_k := \frac{1}{k} - \ln\left(1+\frac{1}{k}\right) > 0
ln⁡(1+1k)=∫kk+1dxx≥1k+12⇒dk≤1k(2k+1)<12k2\ln\left(1+\frac{1}{k}\right) = \int_k^{k+1}\frac{dx}{x} \ge \frac{1}{k+\frac12} \quad \Rightarrow \quad d_k \le \frac{1}{k(2k+1)} < \frac{1}{2k^2}
n=10n = 10: ∑k=110dk=H10−ln⁡11≈0.5310729812\sum_{k=1}^{10} d_k = H_{10} - \ln 11 \approx 0.5310729812, tail <0.05< 0.05 — so 0.5311<γ<0.58110.5311 < \gamma < 0.5811.
Show proof
  1. By problem 2, γ=∑k=1∞dk\gamma = \sum_{k=1}^{\infty} d_k with every dk>0d_k > 0. The first 10 terms telescope to ∑k=110dk=H10−ln⁡11≈2.9289682539−2.3978952728=0.5310729812\sum_{k=1}^{10} d_k = H_{10} - \ln 11 \approx 2.9289682539 - 2.3978952728 = 0.5310729812. Since dk>0d_k > 0, this is already a lower bound: γ>0.5310729812>0.5\gamma > 0.5310729812 > 0.5.
  2. To bound the tail from above, estimate dkd_k from above. Write ln⁡(1+1k)=∫kk+1dxx\ln\left(1+\frac{1}{k}\right) = \int_k^{k+1} \frac{dx}{x}; by convexity of 1x\frac{1}{x}, f(k+t)+f(k+1−t)≥2f(k+12)f(k+t) + f(k+1-t) \ge 2f(k+\tfrac12) for 0≤t≤10 \le t \le 1, so integrating gives ∫kk+1dxx≥1k+1/2=22k+1\int_k^{k+1} \frac{dx}{x} \ge \frac{1}{k+1/2} = \frac{2}{2k+1}. Hence dk≤1k−22k+1=1k(2k+1)<12k2d_k \le \frac{1}{k} - \frac{2}{2k+1} = \frac{1}{k(2k+1)} < \frac{1}{2k^2}.
  3. Now ∑k=11∞dk<12∑k=11∞1k2<12∑k=11∞1k(k−1)=12∑k=11∞(1k−1−1k)=12⋅110=0.05\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 1k2<1k(k−1)\frac{1}{k^2} < \frac{1}{k(k-1)} for k≥2k \ge 2, and the last sum telescopes).
  4. Therefore 0.5310729812<γ<0.5310729812+0.05=0.58107298120.5310729812 < \gamma < 0.5310729812 + 0.05 = 0.5810729812, so certainly 0.5<γ<0.60.5 < \gamma < 0.6. The true value γ≈0.5772\gamma \approx 0.5772 lies inside this interval.

6. Series