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 it were like your proof, the problem wouldn't be fun. –  Troy Woo Aug 23 '14 at 8:22
@TroyWoo It is fun, isn't it? –  Petite Etincelle Aug 23 '14 at 8:26
First,Thank you for you solution,your method is use $e=2.718$.I want know without this approx.because some middle school student don't know this –  math110 Aug 23 '14 at 8:30
@math110 ah ok I see, that't why you ask the question. Hope someone else gives a nice solution –  Petite Etincelle Aug 23 '14 at 8:36
@LiuGang,Yes.That's my mean –  math110 Aug 23 '14 at 8:40

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 $0.971700$

-

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 $1/256$.

-
how to show the remainder is less than $1/256$? –  Petite Etincelle Aug 23 '14 at 8:27
@LiuGang - by the formula for the remainder of the Taylor series. –  nbubis Aug 23 '14 at 19:34

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.

-
I don't know about you, but I certainly couldn't evaluate $\dfrac{\log 8 - \log 3}{\log 17 - \log 16}$ by hand. –  TonyK Aug 23 '14 at 16:08
:) that's true. no simple calculator or log reference probably... –  hypergeometric Aug 23 '14 at 16:15