Proving A_n Converges to 0: Real Analysis

junior33
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prove \ A_n = ( a, a,a,a,a,...) converges to zero. a \in \ R^p

Been reading this real analysis book before i take it next semester and been a lil stuck on this question. I am probably making it seem more difficult than it is. Most of the questions had examples in the chapter but this one didnt. can some one help me out?
 
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If A_n = (1_1,1_2,...1_p) then A_n is constantly point with coordinates 1. This An will NOT converge to 1, it will converge to the constant itself.

i think you are confusing a point in R^p with the sequence itself a point in R^p is a set of p real numbers where the order in which each of these numbers follow each other matters.
a itself is NOT a point in R^p.
 
it says that the a's are vectors in a \in \ R^p

would it be that same
 
sorry, that's correct. the a's ARE vectors. sorry i thought you thought they were coordinates of the vectors in the sequence.

well use the distance function u have and choose N=1 for any epsilon. see what happens.
 
First, as the sticky at the top of this section says, this is NOT the place for homework. I am moving it to the homework section.

Second, you misstated the problem in the body of your post. You do NOT want to prove "that (a, a, a, a, ...) converges to 0" because, in general, it doesn't. You want to prove that it converges to a. Okay what is |a- a|?
 
^^^ yes that's what i meant
 
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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