By how much does (1.0001)^{10000} fall short of e? Give the leading term of the gap as a function of n for (1 + 1/n)^n, and the numerical value at n = 10000.

By how much does (1.0001)^{10000} fall short of e? Give the leading term of the gap as a function of n for (1 + 1/n)^n, and the numerical value at n = 10000.

Approach: Take logarithms, expand log(1+x) to second order in 1/n, then exponentiate and expand once more to isolate the first correction to e.

e/20000 = 1.359e-4. Take logarithms: n log(1 + 1/n) = n (1/n - 1/(2n^2) + 1/(3n^3) - ...) = 1 - 1/(2n) + 1/(3n^2) - ... , using the logarithm expansion. Exponentiating gives (1 + 1/n)^n = e * exp(-1/(2n) + 1/(3n^2) - ...) = e (1 - 1/(2n) + 11/(24 n^2) - ...), so the asymptotic gap is e - (1+1/n)^n = e/(2n) + O(1/n^2), the leading order term. At n = 10000 that is 2.71828/20000 = 1.3591e-4, so (1.0001)^{10000} = 2.718146 against e = 2.718282. The gap shrinks like 1/n, which is why this limit is a poor way to compute e numerically: reaching 10 digits would need n of order 10^{10} and the rounding error in 1 + 1/n would dominate long before that.

Follow-up: What n makes (1 + 1/n)^{n + 1/2} accurate to 10^{-8}, and why does the half shift improve the order?

Key concepts: logarithm expansion, asymptotic gap, exponential expansion, leading order term.