## Calculus, 10th Edition (Anton)

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