Answer
strictly decreasing
Work Step by Step
Given $$\left\{ \frac{10^n}{(2n)!}\right\}_{1}^{\infty}$$
Since $a_n=\dfrac{10^n}{(2n)!},\ \ a_{n+1}= \dfrac{10^{n+1}}{(2n+2)!}$, then
\begin{align*}
\frac{a_{n+1}}{a_n}&=\dfrac{10^{n+1}}{(2n+2)!}\dfrac{(2n)!}{10^ n}\\
&= \dfrac{10}{ (2n+2)(2n+1)}\\
&=\frac{10}{ 4n^2+6n+2} \\
&<1
\end{align*}
So, the sequence is eventually strictly decreasing