# Show that $e^3 > 20$

The inequality $$\boxed{e^3 > 20}$$ is occasionally useful, including in the answer I wrote for this question that comes from a GRE subject exam.

This bound is relatively tight: $$e^3 = 20.08553\!\ldots ,$$ a relative error of $$< \frac{1}{200}$$, which means establishing the inequality might be a little delicate. In a comment under the linked answer, TheSimpliFire posed the following natural question:

What is an efficient way to prove the inequality $$e^3 > 20$$ by hand?

(I would have guessed that this had been asked before, but neither the internal search nor searchonmath turned up any duplicates.)

A naive method is to use the series truncation $$e = \sum_{k = 0}^\infty \frac{1}{k!} > 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{120 + 120 + 60 + 20 + 5 + 1}{160} = \frac{163}{60} .$$ Then, it suffices to prove that $$\left(\frac{163}{60}\right)^3 > 20$$, which is equivalent to $$4\,330\,747 > 4\,320\,000$$. This last step could even be outsourced to an enthusiastic primary school student, but it involves cubing a three-digit prime and so is slightly tedious.

One might try to refine this method by looking for rationals that are easier to cube, but the only rational numbers satisfying $$\sqrt{20} < q < e$$ with denominator $$< 60$$ are $$\frac{106}{39}, \frac{125}{46}, \frac{144}{53}$$. It's again straightforward to show that the cube of any of these $$> 20$$, but doing so is no faster than cubing $$\frac{163}{60}$$ and one then has the additional burden of showing the number is $$< e$$.

One could also search for integrals analogous to the classic Dalzell integrals for the difference $$e^3 - 20$$ (or to the difference corresponding to some other inequality equivalent thereto), by which I mean evidently positive definite integrals equal to that difference.

For example, some experimentation yields the definite integral \begin{align} &\int_1^2 - \frac{(x - 1) (2 - x) p(x) \,dx}{20 x (x^2 + 1)} \\ &\qquad = \int_1^2 \left(-\frac{1}{2} x^3 + \frac{63}{20} x^2 - \frac{153}{20} x + 9 - \frac{3}{x} - \frac{2 x}{x^2 + 1} \right) dx \\ &\qquad = 3 - \log 20 , \end{align} where $$p(x) = 10 x^4 - 33 x^3 + 44 x^2 - 45 x + 30$$. Computing gives that all of the coefficients of $$p(x + 1)$$ are positive, so $$p$$ is strictly positive for $$x \geq 1$$, and thus the integrand is strictly positive on $$(1, 2)$$. So, the integral is positive, that is, $$3 > \log 20$$, which is equivalent via exponentiation to $$e^3 > 20$$. This is again elementary, but not terribly fast.

Remark Incidentally this latter method lets us extract cheap but relatively sharp rational bounds on $$\log 20$$: Since $$2 < x (x^2 + 1) < 10$$ on the interval of integration, our integral is bounded by polynomial integrals: $$\int_1^2 - \frac{(x - 1) (2 - x) p(x) \,dx}{20 \cdot 10} < \int_1^2 - \frac{(x - 1) (2 - x) p(x) \,dx}{20 x (x^2 + 1)} < \int_1^2 - \frac{(x - 1) (2 - x) p(x) \,dx}{20 \cdot 2} .$$ Integrating gives $$\frac{163}{84000} < 3 - \log 20 < \frac{163}{16800},$$ and rearranging gives the bounds $$2.99027\!\ldots = \frac{251185}{84000} < \log 20 < \frac{251837}{84000} = 2.99805\!\ldots .$$

• What about calculating enough terms of the series $e^3 = \sum_{k=0}^\infty \frac{3^k}{k!}$ to exceed 20? Though that does seem to require up to $k=8$. – Daniel Schepler Aug 26 at 17:43
• I think you mean $(\frac{163}{60})^3 \gt 20$ – Ross Millikan Aug 26 at 17:43
• The silliest way is to compute $(2.718)^2 \cdot 2.71$! – dcolazin Aug 26 at 17:49
• Yes, thanks, Ross, fixed! – Travis Willse Aug 26 at 17:52
• @dcolazin That certainly works, of course, but aesthetically I'd prefer methods that use definitions rather than decimal approximations. Of course one can recover $e > 2.718$ as an intermediate step using, for example, the power series for $e^x$ at $x = 1$, but doing so already requires $7$ terms, which is more than the $6$-term calculation I'm looking to improve on. For that matter, $2.718 \cdot 2.718 \cdot 2.71$ certainly requires more single-digit multiplications and carries than $163 \cdot 163 \cdot 163$. – Travis Willse Aug 26 at 18:10

If you know your powers of $$3$$ well, you know $$2.7^3=19.683$$. Since $$e>2.718=2.7\left(1+\frac{2}{300}\right)$$,$$e^3>19.683\left(1+\frac{2}{100}\right)=19.683+0.39366>20.$$

$$1+3+\frac92+\frac92+\frac{27}8+\frac{81}{40}+\frac{81}{80}+\frac{243}{560}+\frac{729}{4480}\\ 13+3.375+2.025+1.025+0.433928\cdots+0.162723\cdots=20.021651$$

isn't so difficult. Only the last two term require a "true" division.

An extended comment.

Not really a proof, but an interesting consequence:

$$\log 20=4 \log 2+\log \left(1+\frac{1}{4}\right)<3$$

$$\log 2< \frac34 -\frac14 \log \left(1+\frac{1}{4}\right)$$

$$\log 2< \frac34 -\frac14 \left(\frac{1}{4}-\frac{1}{32}\right)$$

$$\log 2< \frac34 -\frac1{18}$$

The error here is approximately $$0.0013$$.

That said, there's a lot of inequalities for logarithms, especially for $$\log 2$$ already known. This can be used to prove the OP.

• @Travis, at least $\log 2 < \frac{5^2}{6^2}$ is easy to remember. However, the original inequality doesn't follow from this, because $8^{36/25}<20$. – Yuriy S Aug 26 at 19:11

I am assuming that $$e$$ is defined as the real number satisfying $$\int_1^e \frac{1}{x}=1$$, and that you know $$\int_1^{e^3} \frac{1}{x}dx$$ is $$3$$ and that $$e^3 >20$$ iff $$\int_1^{20}\frac{1}{x} dx$$ is less than 3. And of course that you do not get to use a priori that e.g., $$e \le 2.7189$$

What about evaluating $$\sum_{k=1}^{19} \frac{1}{k}$$ and showing that this is less than 3? This $$\sum_{k=1}^{19} \frac{1}{k}$$ is an upper-bound on $$\int_1^{k+1} \frac{1}{x} dx$$.

• This isn't true, as $\sum_{k = 1}^{19} \frac{1}{k} = 3.54773\,\ldots > 3$. We do have $\log n = H_n - \gamma + \cdots$, but to get a bound on $\log 20$, we need to have some control over the $\cdots$---we can do that using the asymptotic behavior of the Bernoulli numbers, but that's already a significant amount of work. In fact, even just adding the reciprocals $1, \frac{1}{2}, \frac{1}{19}$ is pretty tedious by hand, as $\operatorname{lcm}(1, \ldots, 19) = 232\,792\,560$ is large. – Travis Willse Aug 26 at 18:26
• (And on top of that we'd need to have some way of estimating $\gamma$ rather sharply---which is an interesting question in its own right.) – Travis Willse Aug 26 at 18:40
• (The last sentence of my first comment should be "... $1, \frac{1}{2}, \ldots, \frac{1}{19}$ ...".) – Travis Willse Aug 26 at 20:06
• You could replace this sume with say $\sum_{k=1}^7 \frac{1}{5}(\frac{1}{k}+\frac{1}{k+1/4}+ \frac{1}{k+2/4}+ \frac{1}{k+3/4}+ \frac{1}{k+1})$ $+$ $\sum_{k=8}^{19} \frac{1}{k}$. – Mike Aug 26 at 22:23

How about you simply use the Taylor series for $$e^x$$? $$e^x = \sum_{r=0}^\infty \frac {x^r}{r!} \\ e^3 \gt 1+3/1+9/2+9/2+27/8+ 81/40+81/80+243/560+(243/560)*3/8 + ((243*3)/(560*8))*1/3 = 1+3+9+3.375+2.025+1.0125+0.4339..+0.1627...+0.0542... = 20.0633 \gt 20$$ Basically add up the first ten terms of the Taylor series of e^x with x=3.