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So I right away notice that this $\large\frac{1}{2^n}$ will go to zero, and this can just be replaced by an epsilon.

However, it is not apparent to me how to transition about the statement given about the two sequential elements to any arbitrary element. (if that language is bad, I mean i'm given that $|x_{n+1} - x_n| < \large\frac{1}{2^n}$ $\forall n > N$, and I want to transition that to a statement of $|x_{m} - x_n| < \large\frac{1}{2^n}$ $\forall n > m$. Then the right side just becomes an epsilon as n goes infinity.

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Let $m,n\gt N$, with $m\lt n$. By the Triangle Inequality, we have $$|x_n-x_m|\le |x_{m+1}-x_m| +|x_{m+2}-x_{m+1}| +\cdots +|x_n-x_{n-1}|.$$ The first term on the right is $\lt \frac{1}{2^m}$, the second term is $\lt \frac{1}{2^{m+1}}$, and so on. So the full sum on the right is $\lt \frac{1}{2^{m-1}}$.

For any given $\epsilon\gt 0$, we can find an $N_1\gt N$ such that $\frac{1}{2^{m-1}}\lt \epsilon$.

It follows that if $N_1\lt m\lt n$, then $|a_m-a_n|\lt \epsilon$.

Remark: Informally, for large $m$, we have that $x_{m+1}$ is very close to $x_m$, and $x_{m+2}$ is very close to $x_{m+1}$, and so on. So how far can $x_n$ be from $x_m$? The point is that even in the worst case, if the errors add up, $x_n$ must be close to $x_m$.

Note that if in the problem $\frac{1}{2^n}$ is replaced by $\frac{1}{n}$, the argument breaks down. For the "tail" $\frac{1}{m}+\frac{1}{m+1}+cdots$ of the harmomic series is infinite.

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  • $\begingroup$ thank you, i follow this outside of this part: how did you get the full sum on the right to be $< \large\frac{1}{2^{m-1}}$ ? What about the last part that is $<\large\frac{1}{2^n}$? $\endgroup$ Feb 20 '14 at 4:21
  • $\begingroup$ The sum $\frac{1}{2^m}+\frac{1}{2^{m+1}}+\cdots$ is an infinite geometric series, first term $\frac{1}{2^m}$, common ratio $\frac{1}{2}$. The last term in the sum is reaally $\frac{1}{2^{n-1}}$, but instead of stopping there we added up beyond, getting a bigger number which can still e made small enough. $\endgroup$ Feb 20 '14 at 4:27
  • $\begingroup$ Ah yes, the geometric series - this is just a sequence though, why are we allowed to use an argument from a series? $\endgroup$ Feb 20 '14 at 4:38
  • $\begingroup$ We are adding up bounds on the errors. Adding up means we are using a series. It is a finite series, $m$ to $n-1$. But for an upper bound, we can replace it by an infinite series. The actual estimate we get for the finite sum is $\frac{1}{2^{m-1}}-\frac{1}{2^{n-1}}$, but that's less than $\frac{1}{2^{m-1}}$. $\endgroup$ Feb 20 '14 at 4:41
  • $\begingroup$ that is cool, never actually seen that done before. thank you. $\endgroup$ Feb 20 '14 at 4:44
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Observe that by hypothesis, $$\sum_{k=n}^{\infty} |x_{n+1} - x_{n} | < \frac{1}{2^{n-1}}.$$ Therefore, you can show that $\{x_n\}$ is Cauchy, since given any $\epsilon > 0, \exists N$ so that $\frac{1}{2^{N-1}} < \epsilon$. Therefore, $\forall n,m > N, |x_n - x_m| \le \sum_{n=N}^{\infty} |x_{n+1} - x_n| < \epsilon$. There is a hidden triangle inequality in this last step, but I think you've got it from here.

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Let $m=n+1$. Since $n>N$, then $m>N$. So it becomes $|x_m-x_n|<\epsilon$ (letting $\epsilon=\frac{1}{2^n}$) for $m,n>N$ which is the definition of Cauchy sequence.

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  • $\begingroup$ well, you can let epsilon be that specific value, but the result still needs to hold for all positive epsilon. I see now though that the m we find for each epsilon is just n+1, that makes a lot of sense! $\endgroup$ Feb 20 '14 at 4:02
  • $\begingroup$ You can't "let" $m=n+1$ since the required inequality should hold for all $m,n > N$, not just that one specific value of $m$. $\endgroup$ Feb 20 '14 at 4:48
  • $\begingroup$ If it is true for two consecutive m,n, then it must be true for any m,n since the difference of m,n makes epsilon smaller. $\endgroup$
    – Raven
    Feb 20 '14 at 4:52
  • $\begingroup$ @Rana, I'm not sure what you mean. It seems that by your reasoning a sequence satisfying $|x_{n+1}-x_n| < \frac 1n$ would also be Cauchy, which is not true. $\endgroup$ Feb 20 '14 at 5:48
  • $\begingroup$ @Santiago Canez, I got your point and thanks to point it out......... $\endgroup$
    – Raven
    Feb 20 '14 at 5:52

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