Below is another approach:

The value of $1 invested for n years at interest rate i, compounded x
times per year is (1+(i/x))^{nx}.

In this case f(x) = the value of $1 invested at 100% interest
compounded n times per year = (1+(1/x))^{x}, where x
approaches infinity.

The following table shows values of f(x) for various values of x:

x f(x) ----------- ---------- 1 2.00000000 4 2.44140625 12 2.61303529 52 2.69259695 365 2.71456748 1,000 2.71692393 10,000 2.71814593 100,000 2.71826824 1,000,000 2.71828047 10,000,000 2.71828169 100,000,000 2.71828180

As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.

Michael Shackleford, A.S.A.