The answer is no. Indeed, first of all, to make sense of the question, we need to deal with an infinite sequence of iid $N(0,1)$ random variables (r.v.'s) $X_1,X_2,\dots$. Next, for $n=1,2,\dots,\infty$, let $$Y_n:=\sqrt{\sum_1^n\frac{X_k^2}{k^2}}.$$ Your question can then be stated thus: Is it true that

$$EY_n\underset{n\to\infty}\longrightarrow\sqrt{EY_\infty^2}\,?\tag{1}$$

To answer this question, note first that, by the uniform integrability, $$EY_n\underset{n\to\infty}\longrightarrow EY_\infty.$$ So, the question becomes whether $\sqrt{EY_\infty^2}=EY_\infty$. But the latter equality may occur only if $P(Y_\infty=c)=1$ for some $c\in[0,\infty)$, which implies, in particular, that the r.v. $Y_\infty$ is discrete. In fact, however, the r.v. $Y_\infty$ is clearly absolutely continuous. Thus, the answer to question (1) is indeed no.

The answer is no. Indeed, first of all, to make sense of the question, we need to deal with an infinite sequence of iid $N(0,1)$ random variables (r.v.'s) $X_1,X_2,\dots$. Next, for $n=1,2,\dots,\infty$, let $$Y_n:=\sqrt{\sum_1^n\frac{X_k^2}{k^2}}.$$ Your question can then be stated thus: Is it true that

$$EY_n\underset{n\to\infty}\longrightarrow\sqrt{EY_\infty^2}\,?\tag{1}$$

To answer this question, note first that, by the uniform integrability (see e.g. Corollary 12.8 and Proposition 12.9), $$EY_n\underset{n\to\infty}\longrightarrow EY_\infty.$$ So, the question becomes whether $\sqrt{EY_\infty^2}=EY_\infty$. But the latter equality may occur only if $P(Y_\infty=c)=1$ for some $c\in[0,\infty)$, which implies, in particular, that the r.v. $Y_\infty$ is discrete. In fact, however, the r.v. $Y_\infty$ is clearly absolutely continuous. Thus, the answer to question (1) is indeed no.