Where e Comes From

Question

A bank pays 100% annual interest. Compute your end-of-year wealth on 1 if interest compounds annually, semiannually, quarterly, then derive the limit under continuous compounding, proving which famous constant appears.

Accepted Answer

Compounding n times a year at rate 100% turns 1 into left(1 + frac1nright)^n: n=1: 2, qquad n=2: 2.25, qquad n=4: approx 2.4414, qquad n=12: approx 2.613. The limit. Take logarithms and expand: lnleft(1 + frac1nright)^n = nlnleft(1 + frac1nright) = nleft(frac1n - frac12n^2 + O(n^-3)right) = 1 - frac12n + O(n^-2) to 1. So left(1+frac1nright)^n to e^1 = e approx 2.71828, and the expansion even tells you the convergence speed: the n-fold compounding value is about eleft(1 - frac12nright), which is why n=12 (2.613) still sits noticeably below e. The generalization to make reflexive: left(1 + fracxnright)^n to e^x, and for interview-speed estimates, left(1 - frac1nright)^n approx e^-1 approx 0.368. This single limit powers the matching-hats answer (1/e), bond continuous discounting, Poisson approximations to binomials, and half the "approximately how likely" questions in the bank.

Where e Comes From

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