# Show that the Harmonic series is not cauchy

• ╔(σ_σ)╝
In summary, the conversation discusses a problem involving showing that x_n, defined as the sum of 1/k for k=1 to n, is not Cauchy. One person suggests using algebraic manipulation to prove this, and another suggests finding an epsilon such that for all n_0, there exists m and n greater than or equal to n_0 such that |x_n - x_m| > epsilon. The conversation ends with the conclusion that an epsilon of 1/3 can be found when n = 2m and m >= n_0.
╔(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) .

## 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.

╔(σ_σ)╝ said:

## 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) .

## 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 xn is a sum with n terms, and xm is a sum with m terms, then xn - xm 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?
╔(σ_σ)╝ said:
$$|x_n -x_m| > \frac{m-n}{m+1}$$
╔(σ_σ)╝ said:
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 said:
The right side should be
$$\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n}$$

That was a typo

Mark44 said:
Since xn is a sum with n terms, and xm is a sum with m terms, then xn - xm 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

$$\frac{1}{m+1} + \frac{1}{m+2} +...+ \frac{1}{n} > \frac{n - m}{n} = 1 - m/n$$

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 .

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?

Mark44 said:
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 ?

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

Mark44 said:
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" ?

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

Mark44 said:
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 $$\epsilon >0$$ for all $$n_0$$ such $$m,n \geq n_o$$ then $$| x_n -x_m| > \epsilon$$

I guess ,more rigorously, I could say

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

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 ?

I'm going to turn in, myself.

Okay thanks for the help.

I guess everything is in order.

Off to bed I go. :)

Last edited:

## 1. What is the Harmonic series?

The Harmonic series is an infinite series of the form 1 + 1/2 + 1/3 + 1/4 + ..., where each term is the reciprocal of a natural number.

## 2. What does it mean for a series to be Cauchy?

A series is considered Cauchy if the terms of the series approach 0 as the number of terms approaches infinity. In other words, the terms of the series become closer and closer together as more terms are added.

## 3. Why is it important to show that the Harmonic series is not Cauchy?

Showing that the Harmonic series is not Cauchy is important because it helps us understand the behavior of infinite series and their convergence. It also serves as a counterexample to the common misconception that an infinite number of terms always leads to a finite sum.

## 4. How can we prove that the Harmonic series is not Cauchy?

We can prove that the Harmonic series is not Cauchy by using the Cauchy convergence criterion, which states that for a series to be Cauchy, the sum of the absolute values of the differences between consecutive terms must approach 0 as the number of terms approaches infinity. By evaluating the sum of these differences for the Harmonic series, we can show that it does not approach 0 and therefore the series is not Cauchy.

## 5. What are the implications of the Harmonic series not being Cauchy?

The fact that the Harmonic series is not Cauchy means that it does not converge to a finite sum. This has important implications in mathematics and science, as it shows that not all infinite series behave in the same way and that we must be cautious when dealing with infinite quantities.

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