A result of the exponential formula is $$\exp{\left(\sum_{n=1}^\infty{\frac{a_n}{n!}x^n}\right)}=\sum_{n=0}^\infty{\frac{B_n(a_1,\dots,a_n)}{n!}x^n}$$ Now define $$F(x)=\sum_{n=1}^\infty\frac{1}{(t_n)^x}$$ for some set of real numbers $\{t_1,t_2,\dots\}$.
I can show that for $s\gt0$ $$\exp{\left(\sum_{n=1}^\infty{\frac{-F(ns)} {n}}\right)}=\sum_{n=0}^\infty{\frac{B_n(-0!F(s),-1!F(2s),\dots,-(n-1)!F(ns))}{n!}}$$ where $B_n(x_1,\dots,x_n)$ is the complete exponential Bell polynomial. This equation is what we would get by substituting in the first formula, but without the $x$ variables. But by the same reasoning, we unexpectedly have the following inequality. $$\exp{\left(\sum_{n=1}^\infty{\frac{F(ns)}{n}}\right)}\ne\sum_{n=0}^\infty{\frac{B_n(0!F(s),1!F(2s),\dots,(n-1)!F(ns))}{n!}}$$ I can't figure out how to get an expression for the right hand side that looks somewhat like the left hand side. If there is a way, how do I get it? I'm hoping for a similar relationship as the one with negative terms.