Suppose $X\_n\sim N(0,1) $ is iid, then it is easy to see that
$$\sum\_{n=1}^{\infty}\frac{X\_n}{n}\cos nx$$
converges a.s. for any $x$ since
$$\sum\_{n=1}^{\infty}var(\frac{X\_n}{n}\cos nx)<\infty$$
but how to show that the convergence is uniform? That is, for a mesurable set $A\subset \Omega, \mathbb{P}(A)=1$, the series converges for all $x\in\mathbb{R}$ and $\omega\in A$.



convergence for series of random variables
Let
\begin{equation\*}
 F(x):=\sum\_{n=1}^{\infty}\frac{X\_n}n\,\cos nx. \tag{1}\label{1}
\end{equation\*}
For $j=0,1,\dots$, let
\begin{equation\*}
 s\_j:=\sqrt{\sum\_{2^j\le n<2^{j+1}}E\Big(\frac{X\_n}n\Big)^2}=\sqrt{\sum\_{2^j\le n<2^{j+1}}\frac1{n^2}}. 
\end{equation\*}


For any integer $N>0$ and  

\begin{equation\*}
 S\_N:=\sum\_{N\le n<2N}\frac1{n^2} 
\end{equation\*}
we have $S\_{N+1}-S\_N=-\dfrac1{N^2}+\dfrac1{(2N)^2}+\dfrac1{(2N+1)^2}<0$.
So, $S\_N$ is decreasing in $N=1,2,\dots$ and hence $s\_j$ is decreasing in $j=0,1,\dots$.
Also, $s\_j^2\asymp2^j\frac1{(2^j)^2}\to0$ as $j\to\infty$.


So, by Theorem 1, p. 84 in [Kahane's book](https://www.google.com/books/edition/Some_Random_Series_of_Functions/mquwmHIdT3UC?hl=en&gbpv=1&printsec=frontcover), the function $F$ is continuous almost surely (a.s.). (The condition in that theorem that $s\_j$ be decreasing seems possible to relax.) As noted on p. 48 of Kahane's book, the a.s. continuity of the sum of a random Fourier series of the form \eqref{1} is equivalent to the a.s. uniform convergence of the random series. $\quad\Box$


