I am trying to prove that $$\sum_{n\geqslant 1}\frac{1}{\lfloor e^n\rceil}<5/9$$
Part of this process is to prove that this inequality works without using tedious numerical approximation (stop the sum at a large number), but rather an elegant one.
I tried to separate
$$\lfloor e^n\rceil=f(e^n-1/2)+f(e^n-3/2)+\cdots$$
where
$$f(x)=\frac{1+\text{sign }x}{2}$$
but that got nowhere fast.
Let $$f(n):=\sum_{k=1}^n{1\over\lfloor e^k\rceil}\ .$$ Numerical evidence shows that, e.g., $f(15)\approx0.555053$, and that $f$ does not grow "essentially" after that. I therefore conjecture that you claim is false. For a proof use the estimate $${1\over\lfloor e^n\rceil}-{1\over e^n}={e^n-\lfloor e^n\rceil\over e^n\>\lfloor e^n\rceil}\leq{1/2\over e^n(e^n-1/2)}\leq e^{-2n}\qquad(n\geq1)\ ,$$ and show that the terms with $k\geq 15$ (or $\geq n_0$ for a slightly larger $n_0$) will never bring the sum to $>0.555555\ldots$