Fix $x>0$ and consider the function $$ F(x) := \sum_{n=1}^{\infty} K_{0}(n x) $$ where $K_0$ is the $0^{\mathrm{th}}$-order modified Bessel function of the second kind.
Does this sum diverge? I know that $K_{0}(z) = \frac{e^{-z}}{\sqrt{z}}$ for $z \to \infty$, so it seems feasible that it converges to something. I don't know how to show this.
We know that for $x>0$ and $n\ge1$,
\begin{align}0<K_0(nx)&=\int_0^\infty e^{-nx\cosh(t)}{\rm~d}t\\&<\int_0^\infty e^{-nx(1+t^2/2)}{\rm~d}t\\&=e^{-nx}\int_0^\infty e^{-nxt^2/2}{\rm~d}t\\&\le e^{-nx}\underbrace{\int_0^\infty e^{-xt^2/2}{\rm~d}t}_{{\rm constant~w.r.t.~}n}\end{align}
which is weaker than the inequality you provided.
It remains then to be proven that
$$0<\sum_{n=1}^\infty K_0(nx)<\sum_{n=1}^\infty e^{-nx}\int_0^\infty e^{-xt^2/2}{\rm~d}t$$
converges, which is simply a function of $x$ times a geometric series.
Note the integral converges for any $x>0$. Left up to the reader. $\ddot\smile$