Alternative definition of sequence (real analysis)

StarTiger
Messages
8
Reaction score
1

Homework Statement




Limit of {sn} as n goes to infinity exists provided for all sigma >0 there exists some integer N such that |sn-L| < sigma where n greater than or equal to N.

Prove equivalent to alternate definition:

limit exists provided that for every positive inteer m there exists a real number N such that |sn-L| < 1/m whebever n greater than or equal to N.


The Attempt at a Solution



Well, I know sigma can be anything, so you can replace sigma with 1/m and get Alternative Definition except for N being an integer rather than a real number. I get the idea that N(as fn of sigma) has to be an integer and N(as fn of 1/m) has to be a real number. Actually setting up definition 1 <=> definition 2 is confusing me though.
 
Physics news on Phys.org
This is a pretty useful reformulation and a good exercise in knowing what numbers you are allowed to choose for and which must be arbitrary.

The only difference in the definitions is that 1/m and sigma are switched. Let's say you had the 1/m definition. Since the inequality |s_n - L| < 1/m is already in place, it seems reasonable to require that 1/m < sigma. But in this case, we are given that sigma is an arbitrary positive number, whereas we can choose m since we know the 1/m definition holds (and by Archimedean property). Can you figure out the other direction?
 
Given any sigma> 0, 1/sigma is a positive real number so there exist integer m such that m> 1/sigma whence 1/m< sigma.
 
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...
Back
Top