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?

  • $\begingroup$ Use the Taylor expansion of e^x for x=1. The sum will have positive terms therefore the partial sums will be monotonically increasing $\endgroup$ – Amr May 2 '16 at 13:31
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    $\begingroup$ That is what I did. I'm asking if there is an alternative way that doesn't use the taylor expansion. $\endgroup$ – zz20s May 2 '16 at 13:32
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    $\begingroup$ The Taylor method needs only $\sum_{n=0}^5\frac1{n!}=\frac{163}{60}$, which has a moderate denominator. $163\cdot 7-19\cdot 60 = 1$ $\endgroup$ – Hagen von Eitzen May 2 '16 at 13:38
  • $\begingroup$ @HagenvonEitzen Very true. That doesn't seem too bad. Still, though, I'd curious to know whether there's a different argument. $\endgroup$ – zz20s May 2 '16 at 13:39
  • $\begingroup$ Using $(1+1/n)^n$ needs $n=340$... $\endgroup$ – lhf May 2 '16 at 13:39

$$ \int_{0}^{1} x^2 (1-x)^2 e^{-x}\,dx = 14-\frac{38}{e},$$ 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 $e>\sqrt{\frac{133}{18}}$.

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    $\begingroup$ How did you come across this integral? $\endgroup$ – zz20s May 2 '16 at 13:45
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    $\begingroup$ @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 $\zeta(3)$. $\endgroup$ – Jack D'Aurizio May 2 '16 at 13:46
  • $\begingroup$ Very interesting. This is precisely what I was looking for. $\endgroup$ – zz20s May 2 '16 at 13:47
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    $\begingroup$ $$\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx = \frac{22}{7}-\pi $$ is another classical example. $\endgroup$ – Jack D'Aurizio May 2 '16 at 14:35
  • $\begingroup$ Dear Jack D'Aurizio .. I'm Sorry this comment...Can you look my problem?..Can you show my mistakes?.. I'm grateful to You.. math.stackexchange.com/questions/2678602/… $\endgroup$ – MathLover Mar 6 '18 at 0:37

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.)


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


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



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 $19\times 2$.


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

Compute $7!$ backwards, to get


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

Subtract the terms until you get a negative,


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

  • $\begingroup$ $e>\sum_{j=0}^51/j!= 1+1+.5+.1\bar 6 +.041\bar 6+.008\bar 3= 2.71\bar 6>19/7.$ $\endgroup$ – DanielWainfleet May 2 '16 at 14:26
  • $\begingroup$ @user254665: why do you say this ? $\endgroup$ – Yves Daoust May 2 '16 at 14:27
  • $\begingroup$ It's two terms shorter than your solution. Not that there's anything wrong with yours. $\endgroup$ – DanielWainfleet May 2 '16 at 14:30
  • $\begingroup$ @user254665: no, we use the same terms, up to $5!$ (so $6$ terms); the terms $6!$ and $7!$ aren't used. My purpose is to avoid any division and work with integers only. $\endgroup$ – Yves Daoust May 2 '16 at 14:36

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