Consider $Z_n = \sum_{k=1}^{n} \frac{B_k}{n}$ where each $B_k \sim N(0, k)$ is a Brownian motion. I'm trying to compute the distribution of $Z_n$. Obviously $EZ_n=0$. For $Var Z_n$, since $\mathbf{Cov}B_sB_t = \min\{s,t\},$ I derive \begin{align} Var Z_n &= \frac{1}{n^2}\bigg(\sum_kVar B_k + 2\sum_{i<j}\mathbf{Cov} B_i, B_j\bigg) \\ &= \frac{1}{n^2}\bigg(\binom{n}{2} + 1(n-1) + 2(n-2) + \ldots+(n-1)\bigg) \\ &= \frac{1}{n^2}\bigg(\binom{n}{2} + \binom{n+1}{3}\bigg) \\ &= \frac{1}{2}\bigg(1-\frac{1}{n}\bigg) + \frac{n}{6}\bigg(1-\frac{1}{n^2}\bigg) \end{align} So $Z_n \sim \mathbf{N\bigg(0, \frac{1}{2}\bigg(1-\frac{1}{n}\bigg) + \frac{n}{6}\bigg(1-\frac{1}{n^2}\bigg)\bigg)}$
There are two things that are a bit confusing: I never used the fact that $B_t - B_s \sim N(0, t-s)$ and the sum of dependent normal rvs is not always normal. I'm not sure how to use these facts though.
Linear combinations of jointly normal random variables are always normal. Once you use the fact that $EB_tB_s=\min \{t,s\}$ you don't need the fact that $B_t-B_s \sim N(0,t-s)$.