Today, while browsing Zhihu, I came across a question -
How to elegantly prove $e > \sqrt{7}$
My Solution#
When I saw this question, my initial reaction was to use the series expansion of $e$, $e = \sum^{\infty}_{n=0} \frac{1}{n!}$ to solve it. Of course, I am not a math major, so the following content is not rigorous, just a train of thought.
Expanding it, we get $e = 1+1+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}$,
Taking the first five terms, we calculate and get $e > 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\approx2.7083$,
It is also known that $\sqrt{7} < 2.6458$,
Since $2.7083>2.6458$, we can conclude that $e >\sqrt{7}$.
But after listing out this train of thought, I have to admit that it is not "elegant".
Elegant Solution#
I searched online and found a proof using the Maclaurin series:
According to the Maclaurin series, we know $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$,
Here, we take $x=2$, and we get $e^2 = \sum_{n=0}^{\infty} \frac{2^n}{n!} = 1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \ldots$
Noting that truncating after the first four terms gives us $ \sum_{n=0}^{4} \frac{2^n}{n!} = 7$,
Clearly, $e^2= \sum_{n=0}^{\infty} \frac{2^n}{n!} >\sum_{n=0}^{4} \frac{2^n}{n!} = 7$,
Q.E.D.
Timely Solution#
Because $abcdefg\ldots z$,
thus $e>z$,
According to the Jiang Ping equation, $\text{main}=6$,
and $z=\text{main}-\text{point}=6-0.1=5.9$,
also $5.9^2>7$,
thus $5.9>\sqrt{7}$,
Therefore, $e >\sqrt{7}$,
Q.E.D.
This article is synchronized updated by Mix Space to xLog. The original link is https://ursprung.io/notes/36