# Cauchy Sequence Problem

1. Feb 13, 2007

### lainyg

1. The problem statement, all variables and given/known data

q(n) = Sum(from k=1 to n) 1/n!

Exercise 3: Prove that {q(n)}n(forall)Ns is a cauchy sequence.

2. Relevant equations

none.

3. The attempt at a solution

So many attempts at a solution. I know that a sequence is a cauchy sequence if for all epsilons greater than 0 there exists an N such that m,n >N and therefore the absolute value of q(m) minus (qn) is less than epsilon. A sequence is considered a cauchy sequence of its terms approach a limit (and converge). My problem is with proving this as it is a sum, and not letting it get messy with double factorials. How do I prove this?

2. Feb 13, 2007

### AKG

I think you mean the sum of 1/k!, not 1/n!.

Hint 1: If $\sum _{k = 1} ^{\infty}\frac{1}{k!}$ converges, then for any $\epsilon > 0$, there exists a natural N such that $\sum _{k=N} ^{\infty} \frac{1}{k!} < \epsilon$

Hint 2: What's the Taylor (or Maclaurin) expansion of ex?

3. Feb 14, 2007

### lainyg

OK, so..

if qn converges, then for any epsilon>0 there exists a natural N such that (qn when N=k) is less than epsilon.

With the maclaurin formula we can write that e^x = the sum (from n=0 to infinity) of x^n/n!. Therefore can we just say that since the lim (as n approaches infinity) of q(n) is e, then it converges, and therefore is a cauchy sequence? Or do we still need to show that there's an N such that q(n) is less than epsilon (for any epsilon greater than 0)?

4. Feb 14, 2007

### Dick

I would guess that they are after a more direct proof than just saying 'I know it converges. Thus it is cauchy'. Would it help as a hint to note 1/n!<=1/2^(n) (at least for n>1)?

Last edited: Feb 14, 2007
5. Feb 14, 2007

### lainyg

Another: Let {q(n)n} and {p(n)} (for all integer n's) be Cauchy Sequences which are equivalent. Further let {a(n)} and {b(n)} also be Cauchy Sequences which are equivalent.

Show {q(n) * a(n)} = {p(n) * b(n)} (for all integer n's)