Which is the better approximation to $e$? Let $a_n = (1+1/n)^n$
and $b_n = (1+1/n)^{n+1}$.
Both $a_n \to e$
and $b_n \to e$,
and
$a_n < e < b_n$.
A better approximation to $e$
is known to be
$c_n = (1+1/n)^{n+1/2}
= \sqrt{a_n b_n}
$,
the geometric mean aof $a_n$ and $b_n$.
My question is:
how about $d_n = (a_n+b_n)/2$,
the arithmetic mean of $a_n$ and $b_n$?
Is $d_n$ a better approximation to $e$
than $c_n$?
More precisely,
let $r_n = \frac{c_n-e}{d_n-e}$.
Does $\lim_{n \to \infty} r_n$
exist? If so, is it $0$,
$\infty$, or a finite value?
If the limit is finite, what is it?
Is there some other mean of $a_n$ and $b_n$
(such as the harmonic mean)
which does better than either?
No, I haven't tried to solve these yet.
I thought they would be interesting questions.
Extra points if the answer
does $not$ use the expansions of
$\ln(1\pm x)$ or $e^x$.

We only need third-order approximations to settle this:
$$\ln a_n = n \ln(1 + 1/n) = n(1/n - 1/2n^2 + 1/3n^3 + O(1/n^4)) = 1 - 1/2n + 1/3n^2 + O(1/n^3).$$
$$\ln b_n = (n+1)(1/n - 1/2n^2 + 1/3n^3 + O(1/n^4)) = 1 + 1/2n - 1/6n^2 + O(1/n^3).$$
$$\ln c_n = \tfrac12 (\ln a_n + \ln b_n) = 1 + 1/12n^2 + O(1/n^3).$$
To get asymptotics for $d_n$, expand $a_n = e \exp(\ln a_n - 1) = e( 1 - 1/2n + 11/24n^2 + O(1/n^3))$ and
$b_n = e \exp(\ln b_n - 1) = e(1 + 1/2n - 1/24n^2 + O(1/n^3))$, to see that
$$d_n = e(1 + 5/24n^2 + O(1/n^3)),\text{ while }c_n = e(1 + 1/12n^2 + O(1/n^3)).$$
This explains why $c_n$ is the better approximation by a factor of $2/5$.
An even better approximation would be $(\tfrac12a_n^p + \tfrac12b_n^p)^{1/p}$ for $p=-2/3$, which cancels the second-order term.  Maple calculations show that this in fact also cancels the third-order term and that the coefficient of $n^{-4}$ is quite small (on the order of $1/10000$), so this approximation is surprisingly good!
For $n=100$ we get $2.718281828463159$, good to 10 decimal places and gunning for the 11th.
