Prove Cauchy sequence & find bounds on limit

[*melinda*]

Here's the problem statement:

Prove that $x_1,x_2,x_3,...$ is a Cauchy sequence if it has the property that $|x_k-x_{k-1}|<10^{-k}$ for all $k=2,3,4,...$. If $x_1=2$, what are the bounds on the limit of the sequence?

Someone suggested that I use the triangle inequality as follows:

let $n=m+l$
$$|a_n-a_m|=|a_{m+l}-a_m|$$
$$|a_{m+l}-a_m|\leq |a_{m+l}-a_{m+l-1}|+|a_{m+l-1}-a_{m+l-2}|+...+|a_{m+1}-a_m|$$

Now by hypothesis, $|a_k-a_{k-1}|<10^{-k}$, so

$$|a_{m+l}-a_m|<10^{-(m+l)}+10^{-(m+l-1)}+...+10^{-(m+1)}$$.

It looks like we have an $\epsilon$ such that $|a_n-a_m|<\epsilon$. Before we get to the bounds on the limit, is that correct? Is anything missing?

 

[quantumdude]

You might take it a little further:

$$|a_{m+l}-a_m|<\sum_{i=0}^l 10^{m+l-i}$$

$$|a_n-a_m|<\sum_{i=0}^{n-m} 10^{n-i}$$

I'll let a real mathematician help you the rest of the way.

 

[quicknote]

I would agree with this approach and proof. The use of the triangle inequality is a common method to prove the Cauchy sequence property. Additionally, the use of the given hypothesis is crucial in showing that the sequence is indeed Cauchy.

To find the bounds on the limit of the sequence, we can use the definition of a Cauchy sequence. By definition, a Cauchy sequence is a sequence where for any given positive real number \epsilon, there exists a positive integer N such that for all m,n > N, |a_n-a_m|<\epsilon.

In this case, we can choose \epsilon=10^{-k}, where k is a positive integer. From the proof above, we can see that for any m,n > k, |a_n-a_m|<10^{-(m+l)}+10^{-(m+l-1)}+...+10^{-(m+1)}. Since m and n are both greater than k, we can rewrite this as |a_n-a_m|<10^{-(2k-1)}.

Therefore, the bounds on the limit of the sequence are 10^{-k} for any chosen positive integer k. This means that as k increases, the terms in the sequence get closer and closer together, approaching a limit of 0. This confirms that the sequence is indeed Cauchy.