Is {q(n) * a(n)} = {p(n) * b(n)} (for all integer n's) a Cauchy Sequence?

[lainyg]

Homework Statement

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


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

Homework Equations

none.

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?

 

[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 e^x?

 

[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)?

 

[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)?

 

[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)