Answer
The sequence is strictly decreasing.
Work Step by Step
Given $$\left\{ \frac{5^n}{2n^2}\right\}_{n=1}^{\infty}$$
Since $a_n=\dfrac{5^n}{2n^2},\ \ a_{n+1}= \dfrac{5^{n+1}}{2^{(n+1)^2}}$,
\begin{align*}
\frac{a_{n+1}}{a_n}&= \dfrac{2^{n^2}}{5^n} \times \dfrac{5^{n+1}}{2^{(n+1)^2}}\\=\dfrac{5}{2^{2n+1}} \lt 1
\end{align*}
So, the sequence is strictly decreasing.
