How to show $\displaystyle a_{n}=\frac{4^{2n+1}}{n^{2n}}$ is a descending sequence? I have to show that the sequence
$$a_{n}=\frac{4^{2n+1}}{n^{2n}},\qquad n\geq1$$
is descending.
I thought about proving
$$\frac{a_{n+1}}{a_{n}}<1$$
However, since
$$\frac{a_{n+1}}{a_{n}}=\frac{16n^{2n}}{(n+1)^{2n+2}}$$
I have no clue at all how I could make this.

We have
\begin{eqnarray}
a_1&=&4^3=64\\
a_2&=&\frac{4^5}{2^4}=64
\end{eqnarray}
and for $n\ge 3$,
\begin{eqnarray}
\frac{a_{n+1}}{a_n}&=&\frac{4^{2n+3}\cdot n^{2n}}{(n+1)^{2n+2}\cdot 4^{2n+1}}\\
&=&\frac{16}{(n+1)^2}\cdot\left(\frac{n}{n+1}\right)^{2n}\\
&<&\frac{16}{(3+1)^2}\cdot 1^{2b}\\
&=&1
\end{eqnarray}
i.e. $a_{n+1}/a_n<1$ for $n\ge 3$. Therefore the sequence is decreasing.
