Limit of a Sequence (Epsilon-N)

1. Oct 18, 2009

ksm100

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!

2. Oct 18, 2009

snipez90

I think in this case there is an easier way. 2^n / n! multiplies out as follows: the numerator is just 2*...*2, n times, and the denominator is n*(n-1)*...*2*1. If we divide, we can rewrite as (2/n)*(2/(n-1))*...*(2/2)*(2/1). The terms 2/(n-1), ... 2/2 are each less than 1, which leaves you with (2/n)(2/1) as an upper bound.