Convergence of a series that looks similar to $e^x$

Question

Suppose I have some $\epsilon > 0$ and some constant $c > 0$.

Does the series $$ \sum_{n=1}^{\infty} \frac{c^{n^{\epsilon}} }{[n^{\epsilon}]!}, $$ where $[r]$ is the integral part of a real number $r \in \mathbb{R}$, always converge?

Thanks!

Answer 1

Step $0$: Guess it always converges.

Step $1$: Show that without loss of generality $c>1$.

Step $2$: Show that we only have to show the convergence of $\displaystyle\sum_{n=1}^\infty\frac{c^{\lfloor n^\varepsilon\rfloor}}{\lfloor n^\varepsilon\!\rfloor!}$

Step $3$: Show that $\displaystyle\frac{c^{\lfloor n^\varepsilon\rfloor}}{\lfloor n^\varepsilon\!\rfloor!}$ is nonincreasing for sufficiently high $n$'s.

Step $4$: Use Cauchy's condensation test to get $\displaystyle\sum_{n=0}^\infty2^n\frac{c^{\lfloor(2^\varepsilon)^n\rfloor}}{\lfloor(2^\varepsilon)^n\!\rfloor!}$

Step $5$: Show that $\lfloor(2^\varepsilon)^n\rfloor$ is increasing for sufficiently high $n$'s.

Step $6$: Compare with $\displaystyle\sum_{n=0}^\infty2^n\frac{c^n}{n!}$ and conclude it's convergent.