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

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Homework Statement



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

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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
 
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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?
 
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.
 
If n> N= max(N1, N2), then both n> N1 and n> N2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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