Valid proof that Euler's Constant $\gamma$ is between $0$ and $1$? I was wondering if there is any reasonable way/theory to do calculations with divergent limits of a sequence. I was trying to prove that Euler's constant $$\gamma = \displaystyle{\lim_{n \to \infty}}
\left( \displaystyle\sum_{k=1}^{n} \frac{1}{k} - \log(n)\right)$$ is within $\gamma \in [0,1].$
For me, intuitively it should make sense to rewrite $\gamma$ as
$$\gamma = 
\displaystyle\sum_{k=1}^{\infty} f(x) - \displaystyle{\lim_{n \to \infty}}\log(n)$$
where $f: [1, \infty) \to \mathbb{R}_{\geq0}$ is given by $f(x) = \frac{1}{x}$.
But this step already doesn't seem to work with the theory I learned in Calculus 1. But if we keep going we could now use the integral test for convergence to bound $\gamma$:
$$\gamma \leq 1 + \int_1^\infty \frac{1}{x} dx - \displaystyle{\lim_{n \to \infty}}\log(n)
= 1 + \displaystyle{\lim_{n \to \infty}}\log(n) - \log(1) - \displaystyle{\lim_{n \to \infty}}\log(n) = 1. $$
Although again I see why $\displaystyle{\lim_{n \to \infty}}\log(n) - \displaystyle{\lim_{n \to \infty}}\log(n) = 0$ could be a problematic conclusion. But with the same method we can also establish that
$$\gamma \geq 0.$$
If there were any way to make sense of this idea I think it would be a neat proof, although I'm aware of the problematic conclusions. Maybe anyone can teach me something new or confirm my concerns.
Thanks in advance!
 A: 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
$$
   \gamma_n \le 1 + \int_1^n \frac1x\,dx - \log n = 1 + (\log n - \log 1) - \log n = 1.
$$
A: 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 $1/2\le\gamma\le1$.
