# Show that a sequence is not a Cauchy Sequence

Set $$a_n = 1 + \frac{1}{2}+\ldots +\frac{1}{n}$$ for each $n\in \mathbb N.$ Show that $\{a_n\}$ is not a Cauchy sequence even though $$\lim_{n\to \infty}(a_{n+1}-a_n) =0$$ (Therefore $\{a_n\}$ does not have a limit).

I am really confused on Cauchy sequences but as I tried working through this problem, I think my sequence can be stated as $a_n= 1/n$. So if I use $|a_{n+1}−a_n|$ I get $\frac{1}{n+1} - \frac{1}{n}$ which goes to $0$ as $n$ goes to infinity. And since $1/n$ also goes to $0$ as $n$ goes to infinity, shouldn't that mean that the sequence converges and is a cauchy sequence?

Sorry for my typing, I am still unsure of how to format my questions properly on here. Thanks for any help.

• en.wikipedia.org/wiki/Harmonic_series_(mathematics) – Quantaliinuxite Feb 7 '18 at 0:08
• Hint: For any $n > 0$, $|a_{2n} - a_n| > \frac{1}{2}$. – Daniel Schepler Feb 7 '18 at 0:11
• See the text of my edit for how to use Mathjax to write math on here. Also, images are frowned upon unless necessary, will edit that too. – spaceisdarkgreen Feb 7 '18 at 0:22
• Also: "I think my sequence can be stated as $a_n=\frac{1}{n}.$" No. The problem states what $a_n$ is and it is not that. It is $a_n=1+\frac{1}{2}+\ldots +\frac{1}{n}.$ Thus you have $a_{n+1}-a_n = \frac{1}{n+1},$ which does go to zero. The point of the question is that this does not imply that $a_n$ converges. Daniel Schepler's hint will help you see the sequence is not Cauchy (or equivalently that it is not convergent.) – spaceisdarkgreen Feb 7 '18 at 0:29
• "I think my sequence can be stated as an=1/n." No. Read it again. It specifically states that $a_n = 1 + \frac 12 + \frac 13 + ..... + \frac 1n$. This is NOT $a_n = \frac 1n$. – fleablood Feb 7 '18 at 0:36

In simply terms: A cauchy sequence is one in which after a certain $N$ then ALL possible pairs are close together.

This is a sequence where only pairs of immediately consecutive pairs are close together.

So $|a_n - a_{n+1}|$ might be less than $\epsilon$ and $|a_m - a_{m+1}|$ might be less than $\epsilon$. But $|a_n - a_m|$ might be over a hundred million.

For instance. Take $a_{21} - a_{20}$ that is $\frac 1 {21}$. That's pretty small. But take $a_{40} - a_{20}$ that is $\frac 1{40} + \frac 1{39} + .... + \frac 1{21}$ that is, relatively speaking, pretty large.

We can say, for $\epsilon = \frac 1{20}$ if $n \ge 20$ then $|a_{n+1} - a_n| < \epsilon$. But we can NOT say if $n,m \ge 20$ then $|a_m - a_n| < \epsilon$, because that simply is not true as $m = 40; n = 20$ shows. And it is the LATTER that is the definition of a Cauchy sequence.

In fact consider $a_{1024} - a_{32} = \frac 1{1024} + \frac 1{1023} + ..... + \frac 1{513} + \frac 1{512} + \frac 1 {511}+ ..... + \frac 1{257} + \frac 1{256} + \frac 1{255} + ...... + \frac 1{129}+ \frac 1{128} + \frac 1{127} + .... + \frac 1{65} + \frac 1{64} + \frac 1{63} +.... + \frac 1{33}$

$> \frac 1{1024} + \frac 1{1024} + ..... + \frac 1{1024} + \frac 1{512} + \frac 1{512}+..... + \frac 1{512} + \frac 1{256} + \frac 1{256} + ...... + \frac 1{256}+ \frac 1{128} + \frac 1{128} + .... + \frac 1{128} + \frac 1{64} + \frac 1{64} +.... + \frac 1{34}$

$= 512*\frac 1{1024} + 256*\frac 1{512} + 128*\frac 1{256} + 64*\frac 1{128} + 32*\frac 1{64} = \frac 12 +\frac 12 +\frac 12 +\frac 12 +\frac 12 =2 \frac 12$.

$2\frac 12$ is pretty bigg and we can make this a lot bigger if we needed to.

To be cauchy ANY pair must be close together. And that simply is not the case here.

Proving it is another issue and has a trick. And I gave you a big hint to the trick.

Hint: Prove that $a_{2^k} > \frac k2$.