Which is the better approximation to $e$?

Solution 1:

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.

Solution 2:

This is not an answer but some numerical insight. From the figure, we have $$\lim_{n \to \infty} r_n = 0.4$$ The red line is the line at an height of $0.4$. The limit is almost surely $0.4$, since the value of $r_{50}$ is $\approx 0.39999$. Hence, it is pretty clear that the geometric mean does a better job than the arithmetic mean, but only by a factor. Also, as Ittay Weiss has already mentioned in his comment, the convergence of both the series is actually pretty pretty slow; even for $n=50$, both the sequences, $c_n$ and $d_n$, are only accurate to the third digit.

Similarly, if we were to define $h_n = \dfrac2{\dfrac1{a_n} + \dfrac1{b_n}}$ and look at the sequence $$l_n = \dfrac{h_n-e}{a_n-e}$$ from the figure below we see that $$\lim_{n \to \infty} l_n = -0.2$$ The black line is at a height of $-0.2$. The limit is almost surely $-0.2$, since the value of $l_{50}$ is $\approx -0.19999$. Also, we have $$d_n > c_n > e > h_n$$