Converging sequences (different problem - can you check our thought process?))

mathscott123
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Homework Statement



Suppose that a mathematically inclined child plays with a basket containing an infinite subset of integers (with some repetitions). If an integer k is present in the basket then there are initially |k| copies of it. The child pulls out the integers from the basket at random and arranges them in a sequence ak (a1 is the first integer the child pulls out, a2 is the second, and so on). He/she does not return the integers to the basket. Prove that the sequence bn = 1/an converges. What about a sequence cn = a(n+1) - an.

Homework Equations





The Attempt at a Solution


This is what my group and I came up with for our homework assignment - let us know what you think - thanks!
Given ε>0,choose a natural number M so that 1/M < ε. The set S = {k: |ak|<M} is finite, so max(S) exists. Let n = max(S) + 1. Then for all n > M, an does not belong to S, and hence |an| > M, and |bn|=|1/an| < 1/M < ε. Hence, bn converges to 0.

The sequence cn does not necessarily converge. For example, the basket could have exactly |k| copies of each integer k, and the sequence an could be the sequence of square an = n2. Then cn = 2n + 1, so the sequence diverges.
 
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I think you need to clarify what it means to choose "at random" from an infinite collection of integers. Assuming the collection is countably infinite, there's no way to assign a uniform distribution, so there must be some other distribution. But which one? The answer will depend upon this choice, no?
 
Hmm, I guess the sequence bn = 1/an will converge regardless of what distribution is chosen.

Think about it this way. Given a natural number N, there must be some point in the sequence {an} beyond which there aren't any integers with absolute value less than N. This is because there are only finitely many such integers in the set. Each one either appears in the sequence or it doesn't. If it does, it must appear at some finite index. If it doesn't appear, then it is irrelevant to the problem.

What can you conclude from this?

[edit] - Sorry, I forgot to read your proof. Yes, it looks mostly OK, although be careful that S could be empty if the child didn't choose any small numbers. In that case, max(S) is undefined. But you can simply choose n = 1 in that case. Your counterexample for c_n seems fine.
 
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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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