# Proof on Convergence of Sequence Given Info on Odd/Even Subsequences

1. Nov 13, 2011

### Heute

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

Given that limit of $s_{2n}$ is L and limit of $s_{2n+1}$ is L, prove that lim $s_{n}$ is also L.

2. Relevant equations

3. The attempt at a solution

This seems very obvious: If the even terms of a sequence approach a number and the odd terms of that sequence approach the same number, then the sequence itself approaches that number.

But I'm not sure how to go about translating this into mathematics. I know from the definition of a limit that I can make the odd and even terms of $s_{n}$ as close to L as I want given a large enough n, but what I really need is to go from that to

given e>0 there exists natural number N so that n > N implies |$s_{n}$-L|<e

2. Nov 13, 2011

### Dick

Well, if lim s_2n=L and s_2n+1=L, then that definition of convergence for the odd and evens gives you two N's, right? How about taking the max of the two N's?

3. Nov 14, 2011

### Heute

I thought of that - but I felt like there was still a logical leap from let N = max(N1, N2) to

n > N implies (what we're looking for)

Maybe I'm trying to be too pedantic.

4. Nov 14, 2011

### HallsofIvy

If n> N= max(N1, N2), then both n> N1 and n> N2.