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.
 A: You have$$\frac{a_{n+1}}{a_n}=\left(\frac4{n+1}\right)^2\left(\frac n{n+1}\right)^{2n}.$$It's clear that this is smaller than $1$ if $n>3$. And$$\frac{a_{n+1}}{a_n}=\begin{cases}1&\text{ if }n=1\\\frac{256}{729}&\text{ if }n=2\\\frac{729}{4096}&\text{ if }n=3.\end{cases}$$
A: When $n>4$, then
$$
a_n=\frac{4^{2n+1}}{n^{2n}}=4\left(\frac{4}{n}\right)^{2n}>4\left(\frac{4}{n+1}\right)^{2n}>4\left(\frac{4}{n+1}\right)^{2n+2}=a_{n+1}
$$
And, $a_1=64, a_2=64, a_3=\frac{16384}{729}<23, a_4=4, a_5=\frac{4194304}{9765625}<1$.
A: Hint: Maybe $$\left(\dfrac{n}{n+1}\right)^{2n} = \left(\dfrac{n+1}{n}\right)^{-2n} = \left[\left((1+\dfrac{1}{n}\right)^{n}\right]^{-2}$$ is a known sequence...
A: 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.
A: It is enough to prove that
$$\frac{4^{2(n+1)+1}}{(n+1)^{2n+2}}\le \frac{4^{2n+1}}{n^{2n}}\iff\frac{16}{(n+1)^{2n+2}}\le \frac{1}{n^{2n}}\\16n^{2n}\le(n+1)^{2n+2}\iff 16n^{2n}\le n^{2n}(n+1)^2+(n+1)^2\sum_{k=1}\binom{2n}{k}n^{2n-k}\\16\le(n+1)^2+(n+1)^2\sum_{k=1}\binom{2n}{k}n^{-k}$$
Then one finish verifiant that inequality is true for $n=1$ and $n=2$
A: $$a_{n}=\frac{4^{2n+1}}{n^{2n}}\implies \log(a_n)=(2n+1)\log(4)-2n\log(n)$$
$$\Delta_n=\log(a_{n+1})-\log(a_n)=2 (n \log (n)-(n+1) \log (n+1)+\log (4))$$
Rewrite
$$(n+1) \log (n+1)=(n+1)\log(n)+(n+1)\log \left(1+\frac{1}{n}\right)$$ and use Taylor expansion
$$\Delta_n=\left(\log \left(\frac{16}{n^2}\right)-2\right)-\frac{1}{n}+\frac{1}{3
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1}}{a_n}=e^{\Delta_n}=\frac{16}{e^2 n^2} \Bigg[  1-\frac{1}{n}+\frac{5}{6 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$ So, as soon as $n>\frac 4 e$, that is to say $n\ge 2$, $\frac{a_{n+1}}{a_n}<1$
