Let $\{X_n\}$ be a stationary Gaussian sequence with $EX_0 = 0, EX_0^2 = 1$ and $EX_0X_n = r(n)$. Let $c_n = (2 \ln n)^\frac{1}{2}$ and set $M_n = \max_{0\leqq k \leqq n} X_k$. It is presently known that if $r(n) \ln n = O(1)$, \begin{equation*}\tag{1}\lim \inf\frac{2c_n(M_n - c_n)}{\ln \ln n} = -1 \quad \text{and}\quad \lim \sup\frac{2c_n(M_n - c_n)}{\ln \ln n} = 1\end{equation*} with probability 1. Related results are obtained here assuming $r(n) = o(1)$ and $(r(n) \ln n)^{-1}$ is monotone for large $n$ and $o(1)$. Subject to some regularity in $r(n)$, it is shown that if $r(n) \ln n/(\ln \ln n)^2 = o(1)$, then a.s. \begin{equation*}\tag{2}\lim \inf\frac{2c_n(M_n - (1 - r(n))^{\frac{1}{2}}c_n - Z_n)}{\ln \ln n} = -1 \quad \text{and}\end{equation*} $$\lim \sup\frac{2c_n(M_n - (1 - r(n))^{\frac{1}{2}}c_n - Z_n)}{\ln \ln n} = 1$$ where $Z_n$ is the minimum variance estimate of the mean based on $X_0,\cdots, X_n$. Futhermore if $(\ln \ln n)^2/r(n) \ln n = o(1)$, then a.s. \begin{equation*}\tag{3}\lim_{n\rightarrow\infty} r^{-\frac{1}{2}}(n)(M_n - (1 - r(n))^{\frac{1}{2}}c_n - Z_n) = 0.\end{equation*} It is pointed out that (2) and (3) contain laws for $M_n$ which more closely resemble the one given here in (1). Corresponding results for continuous parameter Gaussian processes are sketched.