$\displaystyle ne^{-\frac{n^2}{8}}$

$\displaystyle ne^{-\frac{n^2}{8}} = \frac{n}{e^\frac{n^2}{8}} $

with quotient rule

$\displaystyle \frac{n}{e^\frac{n^2}{8}}= \frac{(n)'(e^{\frac{n^2}{8}})-(n)(e^{\frac{n^2}{8}})'}{(e^{\frac{n^2}{8}})^2}$

$\displaystyle = \frac{(e^{\frac{n^2}{8}})-(n)(\frac{n}{4})}{(e^{\frac{n^2}{8}})^2}$

$\displaystyle = \frac{1 - \frac{n^2}{4}}{e^\frac{n^2}{8}}$

What can I do next? Or can I assume the function is decreasing now?