I am considering the following problem. Let $(X_j)$ be i.i.d. $N(0,1)$ random variables and $H$ a Hilbert space with orthonormal basis $(e_j)$. Let $$X:=\sum_j \frac{X_j e_j}{j}$$ And for any $\varepsilon>0$, find a compact $K$ so that $\mathbb{P}(X\in K)>1-\varepsilon$.
I have really no idea what to try or why this should work, so any intuition is helpful.
What I gave as a hint was almost right, so let me write out a correction. Fix a constant $c>0$ so that the set $K = \{x: x_j \in [\frac{-c\cdot(\ln(j)^2+1)}{j},\frac{c\cdot(\ln^2(j)+1)}{j}]\}$ is compact (the proof is the same as what I described above: all that matters is that the width is square summable). We would like to show that $P( X \in K)$ can be controlled by $c$. Notice that $$P(X \notin K) = P(\exists j \mbox{ : } |X_j| > c(1 + \ln^2(j) )$$ We can control this with $$\sum_j P(|X_j| > c(1 + \ln^2(j) ) \leq $$ $$\sum_j \frac{2}{\sqrt{2 \pi}}\frac{1}{c(1 + \ln^2(j) )}e^{-c(1 + \ln^2(j) )}$$
Once we notice that $e^{-c\ln^2(j)} = (\frac{1}{j})^{c\ln(j)}$ this sum clearly converges and tends to zero as $c \to \infty$.