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).

Weird Limit

I guess its new puzzle time. I do have a new puzzle to post, but I haven't fully solved it out yet, and I don't like posting puzzles I don't have full solutions to, since I'm worried it will take longer than I expect. Instead I'm going to post a limit I saw a few years ago that I thought was really neat. Thats right, its time for math. Anyway,

lim_N->∞ N sin(N!2πe)

where the limit of N is taken along integer values (which is why I can use factorial instead of gamma or something). Its interesting as the coefficient out front is just going to diverge linearly, and the argument of the sin function is just seemingly random numbers, so it should be random numbers times a huge number, its actually amazing that the limit exists at all.

Anyway, the limit does exist as a nonzero real number, I'll show it next time, but it turns out it converges by the magic of e. Well, not quite just e, if you replaced e by e+r with any rational number r, this limit would still exist (that is a bit of a hint for you).