Alex Miller Alex	Posted on Thursday, 30 January, 2003 - 11:28 pm: Lim N*Sin[2*pi*e*N!] as N tends to infinity. can anyone please help. I must be missing something easy.
Alex Miller Alex	Posted on Friday, 31 January, 2003 - 12:10 am: It goes to 2*Pi but how?
David Loeffler	Posted on Friday, 31 January, 2003 - 09:13 am: Try substituting in the series expansion for $e$, that is ${\displaystyle e=\sum _{r=0}^{\infty }\frac{1}{r!}}$ You should find that you get a lot of integer terms, which can be thrown away as $\sin x=\sin (x+2\pi n)$, a term $1/(n+1)$ and lots of terms that are too small to matter. What you get is thus asymptotically ${\displaystyle N\sin (\frac{2\pi }{N+1}+O(\frac{1}{n^2}))~\frac{2\pi N}{N+1}~2\pi }$ David
Alex Miller Alex	Posted on Friday, 31 January, 2003 - 09:10 pm: thank you very much. -Alex Miller
Alex Miller Alex	Posted on Friday, 31 January, 2003 - 09:14 pm: By the way I really like your site; you have done some beautiful problems.