# Difficult series convergence proof

1. Dec 3, 2011

### The1TL

1. The problem statement, all variables and given/known data
Show that given some ε > 0, there exists a natural number M such that for all n ≥ M, (a^n)/n! < ε

2. Relevant equations

3. The attempt at a solution

Ok so I know this seems similar to a Cauchy sequence problem but its not quite the same. So Im looking for a potential value for M that would complete my proof. So far I have found that if M = ε(a^2+1) then it works, through experimenting with different values for a. However, I cannot seem to find a way to show that (a^(M))/M! < ε. Does anyone know a way to do this? Or should I choose a different value for M?

2. Dec 4, 2011

### Curious3141

Wouldn't it suffice to prove that lim (n→∞) of a^n/n! = 0 for any real positive a? (BTW, that a is assumed to be positive can be inferred from the question statement, since negative a would yield negative values for the term when n is odd, which would violate the ε > 0 stipulation).

This can be done very easily if you're allowed to assume the Taylor series for e^a, which you know converges for all a. This means that the series also satisfies all the tests for convergence, one of which is the limit test (meaning that lim (n→∞) of a^n/n! = 0). The ratio test (ratio between successive terms, which is a/(n+1)) is also met because the limit of the ratio is 0 (which is less than 1).

So you've established that the series consists of positive (when a is positive) terms that steadily decrease, with a limit value of 0. It is therefore assured that if n were set sufficiently high, you would have a term < ε, with subsequent terms continuing to steadily decrease.