Valid proof that Euler's Constant $\gamma$ is between $0$ and $1$?

Solution 1:

What we can do is define $$ \gamma_n = \sum_{k=1}^n \frac1k - \log n. $$ Then $\gamma = \lim_{n \to \infty} \gamma_n$, and (assuming that this limit exists) we can hope to prove $0 \le \gamma \le 1$ by proving that $0 \le \gamma_n \le 1$ for all $n$. This avoids any appearance of the $\infty-\infty$ expressions.

Using the integral approximation, we can put lower and upper bounds on $\gamma_n$. For example, the upper bound you've described becomes $$ \gamma_n \le 1 + \int_1^n \frac1x\,dx - \log n = 1 + (\log n - \log 1) - \log n = 1. $$

Solution 2:

Consider plugging this problem into Euler-Maclaurin summation formula:

$$ \sum_{1\le n\le N}\frac1n=\log N+\underbrace{\frac12-\int_1^\infty{\overline B_1(t)\over t^2}\mathrm dt}_\gamma+\mathcal O\left(\frac1N\right) $$

where $\overline B_1(t)=t-\lfloor t\rfloor-1/2$ is the linear Bernoulli function. using the fact that $|\overline B_1(t)|\le1/2$, we have

$$ \int_1^\infty{|\overline B_1(t)|\over t^2}\le1 $$

This suggest that $0\le\gamma\le1$. This is not the full strength of Euler-Maclaurin summation yet. If we perform integration by parts, sharper bounds can be obtained:

By the fact that $|\overline B_{2n}(t)|\le(-1)^{n-1}B_{2n}$, we have

$$ \int_1^\infty{\overline B_1(t)\over t^2}\mathrm dt=-{B_2\over2}-\int_1^\infty{\overline B_2(t)\over t^3}\mathrm dt\le B_2\left[-\frac12+\int_1^\infty{\mathrm dt\over t^3}\right]=0 $$

This indicates that $1/2\le\gamma\le1$.