How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$

Question

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

Answer 1

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

Answer 2

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 $1/256$.

Answer 3

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 $0.971700$

Answer 4

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.

Answer 5

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} $

Answer 6

$$ (1+\dfrac{1}{16})^{16} = \sum \frac{1}{16^k}\binom{16}{k} = \sum \frac{1}{ 16^k}\frac{16!}{(16-k)!} \frac{1}{k!} < 1 + 1 + \frac{1}{2} + \frac{1}{6} + \dots <\dfrac{8}{3} + \frac{2}{4!}$$

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.

Answer 7

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

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