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 for $Q_2-EQ_2$ and the inequality $E(1/Y^2)<\infty$ for any r.v. $Y$ with a gamma distribution with shape parameter $>2$.

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 for $Q_2-EQ_2$ and the inequality $E(1/Y^2)<\infty$ for any r.v. $Y$ with a gamma distribution with shape parameter $>2$.)

Using the uniform integrability 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. The value of $z_\infty$ is unlikely to be exactly $2$.

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 for $Q_2-EQ_2$ and the inequality $E(1/Y^2)<\infty$ for any r.v. $Y$ with a gamma distribution with shape parameter $>2$.