16

How can I show $\dfrac{19}{7}<e$ without using a calculator and without knowing any digits of $e$?

Using a calculator, it is easy to see that $\frac{19}{7}=2.7142857...$ and $e=2.71828...$

However, how could this be shown in a testing environment where one does not have access to a calculator?

My only thought is to use the Taylor series for $e^x$ with $x=1$ to calculate $\displaystyle e\approx\sum \limits_{n=0}^{7}\frac{1}{n!}=\frac{685}{252}=2.7182...$

However, this method seems very time consuming and tedious, finding common denominators and performing long division. Does there exist a quicker, more elegant way?

Martin Sleziak
  • 53,687
zz20s
  • 6,712
  • Use the Taylor expansion of e^x for x=1. The sum will have positive terms therefore the partial sums will be monotonically increasing – Amr May 02 '16 at 13:31
  • 2
    That is what I did. I'm asking if there is an alternative way that doesn't use the taylor expansion. – zz20s May 02 '16 at 13:32
  • 2
    The Taylor method needs only $\sum_{n=0}^5\frac1{n!}=\frac{163}{60}$, which has a moderate denominator. $163\cdot 7-19\cdot 60 = 1$ – Hagen von Eitzen May 02 '16 at 13:38
  • @HagenvonEitzen Very true. That doesn't seem too bad. Still, though, I'd curious to know whether there's a different argument. – zz20s May 02 '16 at 13:39
  • Using $(1+1/n)^n$ needs $n=340$... – lhf May 02 '16 at 13:39
  • @Bernard I didn't even notice that. Still, the answers this question has generated are pretty interesting, IMO. – zz20s May 02 '16 at 13:57
  • @Bernard No. $\sum_{n=0}^3\frac1{n!}=\frac83<\sum_{n=0}^4\frac1{n!}=\frac{65}{24}<\frac{19}7$ – Hagen von Eitzen May 02 '16 at 19:57
  • @Hagen von Eitzen: Seems like I was out of sorts. I'll delete my comment. – Bernard May 02 '16 at 20:10

5 Answers5

33

$$ \int_{0}^{1} x^2 (1-x)^2 e^{-x}\,dx = 14-\frac{38}{e},$$ but the LHS is the integral of a positive function on $(0,1)$.

Another chance is given by exploiting the great regularity of the continued fraction of $\coth(1)$:

$$\coth(1)=[1;3,5,7,9,11,13,\ldots] =\frac{e^2+1}{e^2-1}$$ gives the stronger inequality $e>\sqrt{\frac{133}{18}}$.

Jack D'Aurizio
  • 353,855
  • 3
    How did you come across this integral? – zz20s May 02 '16 at 13:45
  • 1
    @zz20s: These integrals are commonly used in approximation theory (look for Beuker-type integrals), for instance for estimating the irrationality measure of $\pi,\pi^2$ or $\zeta(3)$. – Jack D'Aurizio May 02 '16 at 13:46
  • Very interesting. This is precisely what I was looking for. – zz20s May 02 '16 at 13:47
  • 4
    $$\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2},dx = \frac{22}{7}-\pi $$ is another classical example. – Jack D'Aurizio May 02 '16 at 14:35
  • Dear Jack D'Aurizio .. I'm Sorry this comment...Can you look my problem?..Can you show my mistakes?.. I'm grateful to You.. https://math.stackexchange.com/questions/2678602/limit-lim-delta-x-to-0-frac-a-delta-x-1-delta-x/2678630#2678630 – MathLover Mar 06 '18 at 00:37
6

The first few convergents of the continued fraction representation of $e$ are $$ 2, 3, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \frac{106}{39} $$

Since these convergents oscillate monotonically towards $e$, the last one works to prove that $e>\frac{19}{7}$.

(If you know that $e$ is irrational, you can stop as soon as you get $\frac{19}{7}$ as a convergent because there will be other terms after that.)

lhf
  • 216,483
1

This is a little easier than the OP's calculation $e\gt\sum_{n=1}^7{1\over n!}={685\over252}\gt{19\over7}$, though not by much: We can show $e^{-1}\lt{7\over19}$ via the truncation of the alternating series

$$e^{-1}\lt1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}+{1\over720}={360-120+30-6+1\over720}={265\over720}={53\over144}$$

and the cross multiplication (with some of the steps retained to make things easy to check by eye)

$$53\cdot19=1060-53=1007\lt1008=980+28=144\cdot7$$

Barry Cipra
  • 79,832
0

Since you know what the result should be, try remove terms one by one by computing residues $r_i$. That will reduce the size of integers, as long as yoou factorize while you can, which is easy because of the factorials (highly composite numbers).

First, notice that the first $3$ terms for $e$ give $\frac{5}{2}$. Now, find residues: $$r_1 = \frac{19}{7} - \frac{5}{2} = \frac{3}{2.7}\,, $$ $$r_2 = \frac{3}{2.7} -\frac{1}{6} = \frac{1}{2}\left(\frac{3}{7}-\frac{1}{3}\right) =\frac{1}{3.7}\,,$$ $$r_3 =\frac{1}{3.7} -\frac{1}{3.8} = \frac{1}{3}\frac{1}{7.8}\,,$$ $$r_4 =\frac{1}{3.7.8} -\frac{1}{120} = \frac{1}{3.8}\left(\frac{1}{7}-\frac{1}{5}\right)\,.$$

Since $5<7$, the last residue $r_4$ is negative, you can stop here, never having to do long multiplications, the hardest being $19\times 2$.

wythagoras
  • 25,026
Laurent Duval
  • 6,452
0

Assume that $8$ terms of Taylor will be enough and estimate $7!\left(\dfrac{19}7-e\right)$.

Compute $7!$ backwards, to get

$$1,7,42,210,840,2520,5040,5040.$$

Initialize with $$7!\cdot\frac{19}{7}=13680.$$

Subtract the terms until you get a negative,

$$8640,3600,1080,240,30,-12.$$

This takes six multiplies, a single division and six subtractions, with integers not exceeding five digits.