4 ms·
You asked WA for the result with 100 digits of precision. It probably is 1 to that precision.
by kidme5 8y ago
You asked WA for the result with 100 digits of precision. It probably is 1 to that precision.
- saagarjha 8y agoI don't think it is. Try N[((1+10^-n)^(10^n))/e, n] for smaller values of n. At first it approaches e faster than the precision can "catch up", and up until around n=50 it will result in 1. But at 10^60 and above it will return 1.0000000…
- btilly 8y agoThere is no need to guess. First of all: (1 + 1/n)^n = e^(log(1 + 1/n) * n) Now from the Taylor series approximation: log(1 + t) = t - t^2/2 + t^3/3 - t^4/4 + ... Substitute that in and we get: e^(log(1 + 1/n) * n) = e^((1/n - 1/(2 n^2) + 1/(3 n^3) - ...) n) = e^(1 - 1/(2n) + 1/(3n^2) - ...) And now you can apply the tangent line approximation to find that this is: e - e/(2n) + O(1/n^2) So if n = 10^N, then the first error should be around the N'th digit.