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!
Best Answer
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 $$ \gamma_n \le 1 + \int_1^n \frac1x\,dx - \log n = 1 + (\log n - \log 1) - \log n = 1. $$