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

  • $\begingroup$ I dont think that with a power of 16, including a transcendent function, make this easy to prove... $\endgroup$ – Masacroso Jul 9 '15 at 21:58

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

  • $\begingroup$ If it were like your proof, the problem wouldn't be fun. $\endgroup$ – Troy Woo Aug 23 '14 at 8:22
  • $\begingroup$ @TroyWoo It is fun, isn't it? $\endgroup$ – Petite Etincelle Aug 23 '14 at 8:26
  • $\begingroup$ 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 $\endgroup$ – math110 Aug 23 '14 at 8:30
  • $\begingroup$ @math110 ah ok I see, that't why you ask the question. Hope someone else gives a nice solution $\endgroup$ – Petite Etincelle Aug 23 '14 at 8:36
  • $\begingroup$ @LiuGang,Yes.That's my mean $\endgroup$ – 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 $0.971700$


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


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


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

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


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

Continuing, we have

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


so, finally,


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.

  • 2
    $\begingroup$ I don't know about you, but I certainly couldn't evaluate $\dfrac{\log 8 - \log 3}{\log 17 - \log 16}$ by hand. $\endgroup$ – TonyK Aug 23 '14 at 16:08
  • $\begingroup$ :) that's true. no simple calculator or log reference probably... $\endgroup$ – hypergeometric Aug 23 '14 at 16:15

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 $1/256$.

  • 3
    $\begingroup$ how to show the remainder is less than $1/256$? $\endgroup$ – Petite Etincelle Aug 23 '14 at 8:27
  • $\begingroup$ @LiuGang - by the formula for the remainder of the Taylor series. $\endgroup$ – nbubis Aug 23 '14 at 19:34

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} $


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

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.

  • $\begingroup$ I'm not sure how you're using the geometric sum on your second line. If $b=(1+\frac1{16})=\frac{17}{16}$ then $1+b+b^2+\cdots=1+\frac{17}{16}+\frac{17^2}{16^2}+\cdots$, rather than what you wrote; and if $b=\frac1{16}$ then $b^{16}-1=\frac1{16^{16}}-1$, not $(1+\frac1{16}^{16})-1$... $\endgroup$ – Steven Stadnicki Jul 9 '15 at 23:02
  • $\begingroup$ @StevenStadnicki thanks I redid it from scratch $\endgroup$ – cactus314 Jul 9 '15 at 23:34
  • 1
    $\begingroup$ You can't '...' here in your second inequality on the top line, since $1+1+\frac12+\frac16=\frac83$ - this is (essentially) equivalent to the standard series for $e$, where the first four terms give the $\frac83$ value but any further terms overshoot. $\endgroup$ – Steven Stadnicki Jul 9 '15 at 23:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.