Find an asymptotic expression of $f(x) = \sum \dfrac{1}{n} \tanh(\frac{x}{n})$ as $x\to \infty$.
What I did: I considered that, for all $n\geq 1$, since $t \mapsto \dfrac{1}{t} \tanh(\frac{x}{t})$ is decreasing for $x\geq0$:
$$\int_{n}^{n+1} \dfrac{1}{t} \tanh(\frac{x}{t})dt\leq\dfrac{1}{n} \tanh(\frac{x}{n})\leq \int_{n-1}^n \dfrac{1}{t} \tanh(\frac{x}{t})dt$$ By substitution ($u = \dfrac{x}{t}$): $$\int_{x/(n+1)}^{x/n} \dfrac{\tanh(u)}{u} du\leq\dfrac{1}{n} \tanh(\frac{x}{n})\leq \int_{x/n}^{x/(n-1)} \dfrac{\tanh(u)}{u} du$$ But then, I don't know what to do. I've tried some integration by parts, but was not able to find anything that converges when I finally sum.
Summing over $n$ your last inequality shows that $f(x)$ goes like $$ \int_0^x \frac{\tanh(u)}{u} \,du \simeq \log(x) $$ since $\frac{\tanh(u)}{u}\simeq\frac1u$.
Moreover, you could prove that $$ \lim_{x\to\infty} [f(x) - \log(x)] $$ exists.