1. The problem statement, all variables and given/known data Determine if a limit exists or if the sequence diverges properly. Justify your answer: the sequence is x_n = 2^n/n! 2. Relevant equations From the definition of a limit, I know that I have to prove that for every epsilon (=E) greater than zero, there exists N (natural number) such that for all n greater than or equal to N, abs value (x_n - x) less than E. 3. The attempt at a solution Assuming the limit exists and is zero, I want to prove that for all E there exists N (natural number) such that for all n >= N, |x_n - x| < E. So I want to show that |2^n/n!| < E. Since 2^n, n! > 0 for all n, 0 < 2^n/n! <= 2^n/n for n>2 so eventually (for n>= 3), 0 < 2^n/n! <= 2(2/3)(2^(n-2)). So if I can show that 2(2/3)(2^(n-2)) < E for large enough N, then I'll prove that the limit exists and is zero. So: (2/3)(2^(n-2)) < E/2 implies (2^(n-2)) < 3E/4 implies (2^n) < 3E so I know I can take the ln of both sides and end up with n ln 2 < ln 3 + ln E n < (ln 3 + ln E)/ ln 2 I know that ln 3 + ln E <= 0 when E < 1/3, which implies that when 0 < E < 1/3, n > ln(3E)/ln(2) but then I'm not sure where to go.. by Archimedean Property, there exists N > ln(3E)/ln(2) which implies that for all n >= N , |x_n - x| < E (when 0 < E < 1/3). But I don't think that's enough, and I'm not sure what to do with the case where E >= 1/3. If anyone could offer any help I would really appreciate it. Thanks!