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
 A: If $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$ then $$16 \log(1+\dfrac{1}{16}) < \log\dfrac{8}{3}$$ Now, let us use a very fast converging series (it contains only positive terms) $$\log\Big(\frac{1+x}{1-x}\Big)=2\sum_{i=0}^{\infty}\frac{x^{2k+1}}{2k+1}$$
and use $x=\frac{1}{33}$. Using only two terms for the summation, we then end (for six exact figures) with $$16 \log(1+\dfrac{1}{16}) \simeq 0.969994 $$ 
Let us do the same with the rhs using $x=\frac{5}{11}$. Using two terms for the expansion already leads to a value of $0.971700$
A: Assuming logs are allowed, and suppose we change the question a little to 
"Find the largest $n$ for which $\left(1+\dfrac 1{16}\right)^n<\dfrac 83$."
The solution would be:
$$\left({\dfrac {17}{16}}\right)^n<\dfrac 83\\
n(\log 17-\log 16)<\log8-\log 3\\
n<\dfrac{\log8-\log 3}{\log 17-\log 16}\\
n<16.18\\
n=16$$
Hence the proposition 
$$\left(1+\dfrac 1{16}\right)^{16}<\dfrac 83$$
is true.
A: \begin{align}
(1+\dfrac{1}{16})^{16} &= \sum_{k=0}^{16} {16 \choose k}(\frac{1}{16})^k \\
& = 2 + \frac{15}{32}  + \frac{35}{256} + \sum_{k=4}^{16} {16 \choose k}(\frac{1}{16})^k \\
& \leq 2 +  \frac{15}{32} + \frac{35}{256} +\sum_{k=4}^{16}  \frac{1}{k!}\\
& \leq 2+  \frac{15}{32} + \frac{35}{256} + e - 1 - 1- \frac{1}{2} - \frac{1}{6}\\
& = e - \frac{2}{3} + \frac{155}{256} \\
& \leq 2.719 - 0.666 + 0.606  = 2.659
\end{align}
I used the fact ${n \choose k} \leq \dfrac{n^k}{k!}$ and $e \geq \sum_{k=0}^{16}\dfrac{1}{k!}$. In addition, $e< 2.719, \frac{2}{3} > 0.666, \frac{155}{256} < 0.606$
Added: for a proof which doesn't use the value of $e$, we could use
\begin{align}
\sum_{k=4}^{16}  \frac{1}{k!} \leq \frac{1}{4!}(1 + \frac{1}{5} + \frac{1}{5\times6} +\frac{10}{5\times 6\times 7}) = \frac{269}{7!} < \frac{39}{6!}<  \frac{7}{5!} = \frac{7}{120} < 0.06
\end{align}
Then we have $2 + \frac{155}{256} + \frac{7}{120} < 2 + 0.606 + 0.06 = 2.666$
A: An easier way would be too look at the series expansion:
$$(1+x)^{1/x}=e- \frac{ex}{2}+O(x^2)$$
Thus,
$$\left(1+\frac{1}{16}\right)^{16}<e-\frac{e}{32}+O(x^2)\approx 2.633 <\frac{8}{3}$$
Where the remainder can be shown to be smaller than $1/256$.
A: 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} $
A: $$ (1+\dfrac{1}{16})^{16}
 = \sum \frac{1}{16^k}\binom{16}{k} 
=  \sum \frac{1}{ 16^k}\frac{16!}{(16-k)!} \frac{1}{k!} < 1 + 1 + \frac{1}{2} + \frac{1}{6} + \dots <\dfrac{8}{3} + \frac{2}{4!}$$
How do we know the remaining terms are small enough?  Let's try an inequality.  In our case, $n=4$.
$$ \sum_{k=n}^\infty \frac{1}{k!} 
 = \frac{1}{n!}\sum_{k=0}^\infty \frac{1}{n^k} = \frac{1}{n!}\frac{1}{1-\frac{1}{n}}< \frac{2}{n!}$$
If we are more careful we can actually prevent the overshooting that @StevenStadnicki points out.
$$1 + 1 + \frac{1}{2}\frac{15}{16} +   \frac{1}{6}\frac{15\times 14}{16\times 16 } + \frac{1}{12}= 2.688 > \frac{8}{3} = 2.66\overline{6}$$
If you extend out to the 4th term the result is 2.654 which is bigger than 2.6379 which is the exact answer up to 4 digits.
