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

\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 $2 + \frac{155}{256} + \frac{7}{120} < 2 + 0.606 + 0.06 = 2.666$

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 $0.971700$

This can be done by hand as a fun little exercise in hexadecimal arithmetic, with some clever up-rounding to keep the calculations from getting too tiresome. Writing everything (including the exponents) in base $16$, with digits $0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F$, the inequality we need to prove can be rewritten as

$$3\cdot11^{10}\lt8\cdot10^{10}$$

Now $11^2=121$ and $121^2=14641$, whether you do the calculations base $10$ or base $16$ (there are no carries in either case). To go further, it helps to use the inequality

$$14641^2\lt14700\cdot14600$$

If this inequality doesn't strike you as obvious (and it shouldn't, really, since we're working in an unfamiliar base), note that

$$14641^2=(14700-BF)(14600+41)=14700\cdot14600-(146\cdot BF-147\cdot41)100-BF\cdot41$$

and

$$147\cdot41\lt200\cdot50=A000\lt B000=100\cdot B0\lt146\cdot BF$$

Continuing, we have

$$11^8=14641^2\lt14700\cdot14600=(147\cdot146)\cdot10^4=1A05A\cdot10^4\lt1A1\cdot10^6$$ and thus

$$11^{10}\lt1A1^2\cdot10^C=2A741\cdot10^C\lt2A800\cdot10^C=2A8\cdot10^E$$

so, finally,

$$3\cdot11^{10}\lt3\cdot2A8\cdot10^E=7F8\cdot10^E\lt800\cdot10^E=8\cdot10^{10}$$

as desired.

Please note, I did all the three-digit multiplications here literally by hand, on paper, so I hope someone will take the time to check my arithmetic and correct it as necessary. The crucial base-$16$ calculations that aren't eyeballable are

$$\begin{align} 121^2&=14641\\ 14641&=14700-BF\\ 147\cdot146&=1A05A\\ 1A1^2&=2A741\\ 3\cdot2A8&=7F8\\ \end{align}$$

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.