TITLE: How can one prove that $e<\pi$?
QUESTION [71 upvotes]: This question is inspired by another one, asking to prove that something approximately equal to $1.2$ is bigger than something approximately equal to $0.9$. The numerical answer to this question was (expectedly) downvoted, though in my opinion it is the most reasonable approach to this kind of problems (${\tiny \text{which I personally find completely useless}}$).
My question will consist of 2 parts:


Prove (without calculator) that $e<\pi$;
Explain what do we learn from the proof/what makes this problem interesting.


Edit: Existing answers only confirm my point of view about various weird inequalitites. Fortunately there is $3$ between $e$ and $\pi$, otherwise the things would be very boring.

REPLY [2 votes]: Use the Leibniz formula for $\pi$:  $$
\frac{\pi}{4} = \sum_{j=0}^\infty \frac{(-1)^j}{2j+1}  \text{.}
$$  This is an alternating sum where each partial sum is alternately an upper and lower bound.  The first few terms constrain $\pi$ to these ranges:
[0,4]
[2.666..., 4]
[2.666..., 3.4666...]
[2.895..., 3.4666...]
[2.895..., 3.33968...]
[2.9760..., 3.33968...]
[2.9760..., 3.28374...]
[3.01707..., 3.28374...]

at which point we know $\pi$ is greater than $3$.
A similar method is this descending series for $\mathrm{e}$:  $$
\mathrm{e} = 3 + \sum_{k=2}^\infty \frac{-1}{k!(k-1)k}  \text{.}
$$
Now after one term, we know $\mathrm{e} < 3-\frac{1}{4} < 3$ and the desired inequality follows.
For determining which of two unknown numbers given only by series is larger, it is helpful to have a series whose partial sums are bounds of some sort.  Since we know we want $\mathrm{e} < \pi$ it is helpful to have a decreasing series for $\mathrm{e}$, an increasing series for $\pi$, and/or alternating series for either or both.
Further lesson: It is very nice to have intervals (or at least bounds) on unfamiliar numbers.  Alternating series are quite handy for this.
Other series for this method: \begin{align}
\mathrm{e}^{-1} &= \sum_{k=0}^\infty \frac{(-1)^k}{k!} &&\text{gives $\mathrm{e}<3$ after $5$ terms} \\
\frac{\pi^2}{8} &= \sum_{k=1}^\infty \frac{1}{(2k-1)^2}  &&\text{gives $3 < \pi$ after 4 terms}
\end{align}
Further further lesson: the more representations of a thing you know, the more you are able to show about it.