Show that the Harmonic series is not cauchy

[╔(σ_σ)╝]

Homework Statement

Let x_n = \sum_{k=1}^{n}\frac{1}{k}

Show x_n is not cauchy.

It seems like a fairly easy problem . I bet my head is just not in the right place tonight ( It's thanksgiving in Canada :D) .

Homework Equations

The Attempt at a Solution

Well I know it is not bounded hence it cannot be cauchy but I doubt I am supposed to use this. I guess I am supposed to "show" it by some sort of algebraic manipulation.For n > m
|x_n - x_m| = \frac{1}{m+1} + \frac{1}{m+1} +...+ \frac{1}{n}

|x_n -x_m| > \frac{m-n}{m+1}I am trying to show that I can find an \epsilon >0 for all n_0 such m,n \geq n_o then | x_n -x_m| > \epsilon

I am kinda stuck at this point. :(

I want to find some sort of \epsilon in terms of n_0 but I am having a hard time.

I can find \epsilon in terms of n,m but obviously that is not useful to me.

So I am looking for a way to relate \frac{m-n}{m+1} to some inequality involving n_0

Any hints would be good.

 

[Mark44]

Homework Statement

Let x_n = \sum_{k=1}^{n}\frac{1}{k}

Show x_n is not cauchy.

It seems like a fairly easy problem . I bet my head is just not in the right place tonight ( It's thanksgiving in Canada :D) .

Homework Equations

The Attempt at a Solution

Well I know it is not bounded hence it cannot be cauchy but I doubt I am supposed to use this. I guess I am supposed to "show" it by some sort of algebraic manipulation.


For n > m
|x_n - x_m| = \frac{1}{m+1} + \frac{1}{m+1} +...+ \frac{1}{n}

The right side should be
\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n}

Since x_n is a sum with n terms, and x_m is a sum with m terms, then x_n - x_m has n - m terms. The sum above is larger than (n - m) times the smallest term, or
\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n} > \frac{n - m}{n} = 1 - m/n
Can you do anything with that?

|x_n -x_m| > \frac{m-n}{m+1}

I am trying to show that I can find an \epsilon >0 for all n_0 such m,n \geq n_o then | x_n -x_m| > \epsilon

I am kinda stuck at this point. :(

I want to find some sort of \epsilon in terms of n_0 but I am having a hard time.

I can find \epsilon in terms of n,m but obviously that is not useful to me.

So I am looking for a way to relate \frac{m-n}{m+1} to some inequality involving n_0

Any hints would be good.

 

[╔(σ_σ)╝]

The right side should be
\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n}

That was a typo

Since x_n is a sum with n terms, and x_m is a sum with m terms, then x_n - x_m has n - m terms. The sum above is larger than (n - m) times the smallest term, or
\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n} > \frac{n - m}{n} = 1 - m/n
Can you do anything with that?

XD. I am really not thinking today. I wrote down

<br /> \frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n} &gt; \frac{n - m}{n} = 1 - m/n<br />

in my notes but wrote the oppsite on the thread :( This is actually the point where I got stuck at.I want to find an epsilon such that

1- m/n > epsilon .

 

[Mark44]

Epsilon is generally pretty small. You can reasonably assume that epsilon will be smaller than, say 1/2. Can you fiddle with m and n to make 1 - m/n > 1/2?

Given an arbitray (small) epsilon, can you fiddle with m and n to make 1 - m/n > epsilon?

 

[╔(σ_σ)╝]

Epsilon is generally pretty small. You can reasonably assume that epsilon will be smaller than, say 1/2. Can you fiddle with m and n to make 1 - m/n > 1/2?

Given an arbitray (small) epsilon, can you fiddle with m and n to make 1 - m/n > epsilon?

Sure I could make

n > 2m.

Hmm... I guess that could work.

So in summary if I can make n > 2m I can pick \epsilon = \frac{1}{2} which gives me my desired result.

I need some sleep; I am losing common sense at this point.


Is what I wrote down correct and error free ?

 

[Mark44]

Yeah, that works. Think about how you would make 1 - m/n > epsilon.

 

[╔(σ_σ)╝]

Yeah, that works. Think about how you would make 1 - m/n > epsilon.

Wait, this is a new question ?

Can you explain a bit more what you are asking ? :)

You want me to come up with a "formula" ?

 

[Mark44]

To show divergence, you need to negate the definition for Cauchy convergence.

 

[╔(σ_σ)╝]

To show divergence, you need to negate the definition for Cauchy convergence.

I already "did" in my original post ; however, is was not very formal.

I am trying to show that I can find an \epsilon &gt;0 for all n_0 such m,n \geq n_o then | x_n -x_m| &gt; \epsilon

I guess ,more rigorously, I could say

There exist \epsilon &gt;0 such that for all n_0 in N there exist m,n \geq n_o such that | x_n -x_m| &gt; \epsilon

 

[╔(σ_σ)╝]

As already shown if

n = 2m

m >= n_0

and epsilon is 1/3

I have already satisfied my definition since I have found an epsilon and both n and m.
Did I make a error ?

 

[Mark44]

I'm going to turn in, myself.

 

[╔(σ_σ)╝]

Okay thanks for the help.

I guess everything is in order.

I appreciate your patience.

Off to bed I go. :)