For $r>0$ we define $$\operatorname{E}_1(r) := \int_{r}^\infty \frac{e^{-t}}{t}dt.$$ This defines a positive strictly decreasing function with logarithmic singularity at $r=0$. Breaking the integral at $t=1$ gives you the estimate $$\operatorname{E}_1(r) \le \max(0,-\log(r))+e^{-r}.$$ I need to estimate $$f(r):=\sum_{n =1}^\infty \operatorname{E}_1(n^2r).$$ Is there a $C>1$ with $$f(r) \le C \operatorname{E}_1(r)$$ for all $r>0$? If so, what is the least such $C$?
Or do we have $$ f(r)<-C\log(r) $$ for some $C>1$ for small values of $r$?
Obviously, $$f(r)=\int^\infty_r\sum^\infty_{n=1}e^{-n^2t}\,\frac{dt}t.\tag1$$ But $$\lim_{h\to0}\sum^\infty_{n=1}h\,e^{-n^2h^2}=\int^\infty_0e^{-x^2}\,dx=\frac{\sqrt{\pi}}2,$$ with $h=\sqrt{t}$ we get $$\sum^\infty_{n=1}e^{-n^2t}\sim\frac12\,\sqrt{\frac{\pi}t}.$$ Plugging this into (1), we have $$f(r)\sim\sqrt{\frac{\pi}r}.$$