infinite sum of normal r.v. is still a normal r.v. when given $\sum \limits_{i=1}^\infty a_i^{2}$ is finite If $X_1, X_2, ...$ are i.i.d.standard normal random variables and for real constants $a_1, a_2, ...$, given $\sum \limits_{i=1}^\infty a_i^{2} $ is finite, then $Y_n =\sum\limits_{i=1}^n a_iX_i$ converges (to $Y=\sum a_iX_i$) in $L^{2}(\Omega, P)$ and Y is also a normal random variable.
I don't know how to show Y converges in $L^{2}(\Omega, P)$, but if anyone can prove this, then I can show Y is normal.
I try to show that $E(Y_n-Y)^{2} \rightarrow 0$, but can't do it.
 A: Let $m < n \in \mathbb{N}$. Then, by the independence of the random variables (Bienaymé formula),
$$\begin{align*} \left\| \sum_{i=1}^n a_i X_i - \sum_{i=1}^m a_i X_i \right\|_{L^2}^2 &= \left\| \sum_{i=m+1}^n a_i X_i \right\|_{L^2}^2 \\ &= \sum_{i=m+1}^n a_i^2 \underbrace{\|X_i\|_{L^2}^2}_{\text{var}(X_i)=1} \\ &\stackrel{n,m \to \infty}{\to} 0, \tag{1}\end{align*}$$
i.e. $\left( \sum_{i=1}^n a_i X_i \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $L^2(\mathbb{P})$. Consequently, there exists
$$Y:=\sum_{i=1}^{\infty} a_i X_i := L^2-\lim_{n \to \infty} \sum_{i=1}^n a_i X_i.$$ 
A: We can show that the sequence $Y_n=\sum_{i=1}^na_iX_i$ is a Cauchy sequence. Suppose that $n>m$. Then
$$
\operatorname E[Y_n-Y_m]^2=\operatorname E\bigg[\sum_{i=m+1}^na_iX_i\biggr]^2=\operatorname EX_1^2\sum_{i=m+1}^na_i^2
$$
using the independence and identical distributions of $X_1,X_2,\ldots$. Since the series $\sum_{i=1}^\infty a_i^2$ converges, the term $\sum_{i=m+1}^na_i^2$ can be made arbitrarily small. The space $L^2(\Omega,P)$ is a complete metric space. Hence, sequence $Y_n$ converges since it is a Cauchy sequence.
