I am attempting to solve a problem which requires finding $\lim_{n\rightarrow\infty} \sum_{k=1}^n \lceil e^ek+\frac{1}{\sqrt{5} k} \rceil - \lceil e^ek-\frac{1}{\sqrt{5} k} \rceil$.
My attempt at solving this problem went as follows:
Given that we know $n \leq \lceil n \rceil <n+1 $ and $m \leq \lceil m \rceil <m+1 $ where n and m are real numbers, $\lceil n \rceil - \lceil m \rceil$ can be described as $n-m -1 \leq \lceil n \rceil - \lceil m \rceil < n-m +1 $. Applying this to the original equation and symplifying a bit, we know:
$\frac{2}{\sqrt{5} k} -1 \leq \lceil e^ek+\frac{1}{\sqrt{5} k} \rceil - \lceil e^ek-\frac{1}{\sqrt{5} k} \rceil < \frac{2}{\sqrt{5} k}+1 $
Next I took the infinite sum of the inequality:
$\sum_{k=1}^\infty (\frac{2}{\sqrt{5} k}) - \sum_{k=1}^\infty (1) \leq \sum_{k=1}^\infty ( \lceil e^ek+\frac{1}{\sqrt{5} k} \rceil - \lceil e^ek-\frac{1}{\sqrt{5} k} \rceil) < \sum_{k=1}^\infty (\frac{2}{\sqrt{5} k}) + \sum_{k=1}^\infty (1) $
The right side of the inequality clearly diverges to $\infty$ because $\sum_{k=1}^\infty (\frac{2}{\sqrt{5} k}) = \frac{2}{\sqrt{5}} \times$ (the harmonic series), and that added to another divergent series must diverge.
As for the left side of the inequality, I am unsure how to mathematically show that the sequence diverges to $-\infty$, although I was able to confirm its divergence using Wolfram Alfa.
I am fairly sure the orgininal problem diverges to $\infty$ but using my method I was only able to confirm that the solution is between $-\infty$ and $\infty$. I was wondering if it is possible to manipulate the left side of the inequality to show it diverges or if there is any other convergence test or methods to solve this problem.