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No. The conjugate of the constant zero is the indicator of zero (and vice versa). More generally, the conjugate of the indicator of a closed convex set is positively one-homgeneous (and hence, not super coercive).

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According to this answer, $$z_\infty:=\lim_n z_n=I:=\int_0^\infty F(s)G(s)\,ds,$$ where $$F(s):=\prod_{k=1}^\infty\frac{1}{\sqrt{1+2s/k^3}},\quad G(s):=\sum_{k=1}^\infty\frac k{k^3+2s}.$$ Mathematica can express $F$ and $G$ in terms of functions built-in in Mathematica (and these expressions should be rather straightforward to verify), and then the ...

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Using the uniform integrability of what is under your expectation sign, we have $$z_\infty:=\lim_n z_n=E(Q_2/Q_3),$$ where $$Q_p:=\sum_1^\infty\frac{Z_k^2}{k^p}$$ and the $Z_k$'s are independent standard normal random variables (r.v.'s). The value of $z_\infty$ is unlikely to be exactly $2$. (The uniform integrability follows, say, by Rosenthal's inequality ...

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