$(4pts)2b_n=$

$(n-5)e^{n}is$ a decreasing sequence.

Look $at$ the first few terms $of{b}_n$ :

$b_1=-4e^1$~~$-10\mathrm{.}87$

$b_2=-3e^2$~~$-22\mathrm{.}67$

$b_3=-2e^3$~~$-40\mathrm{.}17$

Since $b_1>b_2>b_3>$dots then $b_{n}is$ a decreasing sequence.