If $f(x)=x$ for $0 \leq x \leq e$ and $f(x)=xf(\ln(x))$ for $x>e$, then does the series $\sum\limits_n\frac1{f(n)}$ converge?

The function $f$ is increasing hence $$ \sum_{n\geqslant1}\frac1{f(n)}\geqslant\sum_{n\geqslant1}\int_n^{n+1}\frac{\mathrm dx}{f(x)}=\int_1^\infty\frac{\mathrm dx}{f(x)}. $$ Define a sequence $(x_n)$ by $x_0=1$ and, for every $n\geqslant0$, $x_{n+1}=\mathrm e^{x_n}$, and, for every $n\geqslant0$, let $$ J_n=\int_{x_n}^{x_{n+1}}\frac{\mathrm dx}{f(x)}. $$ Then, for every $n\geqslant1$, the change of variable $t=\ln x$ sends the interval $(x_n,x_{n+1})$ to the interval $(x_{n-1},x_n)$ hence $$ J_n=\int_{x_n}^{x_{n+1}}\frac{\mathrm dx}{xf(\ln x)}=\int_{x_{n-1}}^{x_{n}}\frac{\mathrm dt}{f(t)}=J_{n-1}. $$ Thus, for every $n\geqslant0$, $$ J_n=J_0=\int_1^\mathrm e\frac{\mathrm dx}x=1, $$ in particular, $$ \int_1^\infty\frac{\mathrm dx}{f(x)}=\sum_{n\geqslant0}J_n $$ diverges, hence the series $\sum\limits_{n\geqslant1}\frac1{f(n)}$ diverges.