Weird Answer

Alright, time to post the solution to the limit from last time. To start with, let us consider the following function

f(x)=sin(2πx)

for our puzzle, we have to worry about x=N!e. Clearly f is periodic in x with period 1, and so for this function only the non-integer part of x matters. That is f(2.7)=f(0.7), f(5749.276)=f(0.276), and f(x+N)=f(x) for integer N. Now consider

lim_N->∞f(N!r)

for N going to infinity along the integers and r any rational number. Since r is rational, r=p/q for p and q integers. Once N is greater than q we will have that N!r will always be an integer, so the limit will necessarily be zero. This proves that

lim_N->∞f(N!(e+r))=lim_N->∞f(N!e+N!r) =lim_N->∞f(N!e)

Proving the statement I made last time. Of course, since e is irrational, we must be a bit more careful. To proceed, we must use the following formula for e

e=Σ1/k!

Summing k from 0 to infinity. If you don't recall this formula, you can derive it from the taylor expansion for e^x and sub in x=1.

Anyway, so N!e then looks like

N!e=ΣN!/k! =N!/0!+N!/1!+N!/2!+...N!/N!+N!/(N+1)!+N!/(N+2)!+... =M+1/(N+1)+1/(N+1)(N+2)+...

For M being some integer.

So, we must have

f(N!e)=f(M+1/(N+1)+1/(N+1)(N+2)+...) =f(1/(N+1)+O(1/N²))

The O(1/N²) means terms like 1/N² or smaller (as N is going to be large). We know that sin(x+O(x²))=x+O(x²) for small x though, so we must have

f(N!e)=2π/(N+1)+O(1/N²)

Which means that

lim_N->∞ N sin(N!2πe) =lim_N->∞ 2πN/(N+1)+O(1/N²) =2π

Ending the problem.

Essentially the proof comes down to the fact that N!e gets arbitrarily close to integer values for large N (as we have seen, it gets within 1/N of an integer). Actually, if you try to do this limit on any computer it almost certainly won't work, as the computers approximate value of e will explode terribly when multiplied by N!. I recall that wolfram alpha can't handle this limit (or at least I have never found a way to coax it into getting this limit).