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.