<s>[INST] How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$ 
show  that
  $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$

it's well know that 
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand
 [/INST] We can split the series at index $m \in \mathbb{Z}$ where $0 \le m \le 15$: 
$S = \Big(1+\dfrac{1}{16}\Big)^{16} = \displaystyle\sum\limits_{k=0}^{m}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} + \displaystyle\sum\limits_{k=m+1}^{16}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} \tag{1}$
Now let $a_k$ be the terms in the summation, and find the ratio between successive terms, so
$\dfrac{a_{k+1}}{a_k} = \dfrac{16!}{(k+1)!(16-k-1)!}\cdot\dfrac{1}{16^{k+1}}\cdot\dfrac{k!(16-k)!}{16!}\cdot 16^k = \dfrac{16-k}{k+1}\cdot\dfrac{1}{16}$
So for further ratios
$\dfrac{a_{k+p+1}}{a_{k+p}} = \dfrac{16-k-p}{k+p+1}\cdot\dfrac{1}{16} \le \dfrac{16-k}{k+1}\cdot\dfrac{1}{16}\ \ \text{ for }p \ge 0 $
and then $\dfrac{a_{m+p}}{a_m} \le \Big(\dfrac{16-m}{16(m+1)}\Big)^p\ \ (\forall p \in \mathbb{Z}_{\ge0}) \tag{2}$
Hence the last term in (1) can be bounded as
$\omega = \displaystyle\sum\limits_{k=m+1}^{16}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} \le \displaystyle\sum\limits_{p=1}^{16-m}{\Big(\dfrac{16-m}{16(m+1)}\Big)^p{a_m}}$
To avoid the upper limit of summation, compare with a sum to infinity:
$\omega \le a_m\displaystyle\sum\limits_{p=1}^{\infty}{\Big(\dfrac{16-m}{16(m+1)}\Big)^p} = \dfrac{16-m}{17m}a_m \tag{3}$
So write $S$ as a sum of terms from $k=0$ to $k=m-1$ and a term encompassing $a_m$ and $\omega$:
$S = \displaystyle\sum\limits_{k=0}^{m-1}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} + a_m\Big(1+\dfrac{\omega}{a_m}\Big) \le \Bigg[\displaystyle\sum\limits_{k=0}^{m-1}{{16 \choose k}\Big(\frac{1}{16}\Big)^k}\Bigg] + \dfrac{16}{17}\Big(1+\dfrac{1}{m}\Big){16 \choose m}\Big(\frac{1}{16}\Big)^m$
For $m=2$, we get
$S \le 1 + 1 + \dfrac{16}{17}\Big(1+\dfrac{1}{2}\Big){\dfrac{16\times15}{1\times2}}\Big(\dfrac{1}{16}\Big)^2 = 2 + \dfrac{45}{68} < 2 + \dfrac{2}{3} $
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