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Why is this a valid proof for the harmonic series?

Why do we say the harmonic series is divergent?improper integrals: Need help pleaseWhat's the use of this theorem about series?Necessary condition for the convergence of an improper integral.How to show that $intlimits_1^{infty} frac{1}{x}dx$ diverges(without using the harmonic series)?Proof that the improper integral $frac{1}{(x^2-1)}$ from $0$ to $infty$ is divergent.A property of the _digamma_ functionproof of integral divergingValid Series Convergence Proof?Is there an intuitive reason as to why the harmonic series is divergent?

1

1

$begingroup$

The other day, I encountered the Harmonic series which I though is an interesting concept. There were many ways to prove that it's divergent and the one I really liked was rather simple but did the job nevertheless. But a proof used integrals and proved that it diverges. But when using integrals, aren't we also calculating fractions with a decimal denominator? I'm not really familiar with calculus and usually only do number theory but doesn't integral find the area under a continuous curve?
Thanks in advance for any help!
EDIT Here's the integral proof:
$$int_1^{infty} 1 / x ,dx= {infty}$$

calculus integration divergent-series

share|cite|improve this question

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It would really help us if you included the proof in question...
  $endgroup$
  – Eevee Trainer
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer my question isn't about a proof for the Harmonic series. I know it diverges using comparison test and fully understand it. But why does the integral work? Here's the comparison test: en.wikipedia.org/wiki/…. EDIT I just realized you meant the integral proof! I'm so sorry, I'll add it.
  $endgroup$
  – Borna Ghahnoosh
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your textbook probably has the "integral test" ... look in the index for that.
  $endgroup$
  – GEdgar
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because the function is decreasing so for example on $[1,2]$, $1 ge frac{1}{x}$, on $[2,3]$, $frac{1}{2} ge frac{1}{x}$ etc; integrating and summing on $n$ we get the result
  $endgroup$
  – Conrad
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

The other day, I encountered the Harmonic series which I though is an interesting concept. There were many ways to prove that it's divergent and the one I really liked was rather simple but did the job nevertheless. But a proof used integrals and proved that it diverges. But when using integrals, aren't we also calculating fractions with a decimal denominator? I'm not really familiar with calculus and usually only do number theory but doesn't integral find the area under a continuous curve?
Thanks in advance for any help!
EDIT Here's the integral proof:
$$int_1^{infty} 1 / x ,dx= {infty}$$

calculus integration divergent-series

share|cite|improve this question

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It would really help us if you included the proof in question...
  $endgroup$
  – Eevee Trainer
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer my question isn't about a proof for the Harmonic series. I know it diverges using comparison test and fully understand it. But why does the integral work? Here's the comparison test: en.wikipedia.org/wiki/…. EDIT I just realized you meant the integral proof! I'm so sorry, I'll add it.
  $endgroup$
  – Borna Ghahnoosh
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your textbook probably has the "integral test" ... look in the index for that.
  $endgroup$
  – GEdgar
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because the function is decreasing so for example on $[1,2]$, $1 ge frac{1}{x}$, on $[2,3]$, $frac{1}{2} ge frac{1}{x}$ etc; integrating and summing on $n$ we get the result
  $endgroup$
  – Conrad
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The other day, I encountered the Harmonic series which I though is an interesting concept. There were many ways to prove that it's divergent and the one I really liked was rather simple but did the job nevertheless. But a proof used integrals and proved that it diverges. But when using integrals, aren't we also calculating fractions with a decimal denominator? I'm not really familiar with calculus and usually only do number theory but doesn't integral find the area under a continuous curve?
Thanks in advance for any help!
EDIT Here's the integral proof:
$$int_1^{infty} 1 / x ,dx= {infty}$$

calculus integration divergent-series

share|cite|improve this question

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

$endgroup$

share|cite|improve this question

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

share|cite|improve this question

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

edited 3 hours ago

Michael Rybkin

4,9441524

edited 3 hours ago

Michael Rybkin

4,9441524

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

asked 4 hours ago

Borna GhahnooshBorna Ghahnoosh

83

2 Answers
2

active

oldest

votes

3

$begingroup$

Consider the following summation:

$$
A=1cdot 1+frac12cdot1+frac13cdot1+frac14cdot1+...
$$

That's a sum of an infinite number of rectangles of height $frac1n$, where $ninmathbb{N}$, and width 1. Do you agree that's nothing more than the harmonic series? Moreover, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ is clearly less than $A$ and lies totally within $A$ (see the picture below). However, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ doesn't add up to a finite number. It goes to infinity. What should be happening with a piece of area that's even larger than that? Obviously, it must also be infinite. Therefore, the harmonic series diverges.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

$endgroup$

add a comment | 

4

$begingroup$

It works using a simple inequality. If $f(x) leq M$ for $x in [a,b]$, then $int_a^b f(x),dx leq M(b-a)$. This is fairly easy to prove:

$$int_a^b f(x),dx leq int_a^b M,dx = M(b-a).$$

Notice that $1/x < 1/n$ on the interval $[n,n+1]$. Then,

$$infty = int_1^infty frac{1}{x}dx = sum_{k=1}^infty int_k^{k+1} frac{1}{x},dx leq sum_{k=1}^infty int_k^{k+1}frac{1}{k},dx = sum_{k=1}^infty frac{1}{k}.$$

share|cite|improve this answer

answered 4 hours ago

forgottenarrowforgottenarrow

33617

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The theorem as you state it is valid only for finite sums. However, you use it for an infinite sum in your explanation. As I'm sure you're aware, this doesn't always follow.
  $endgroup$
  – Allawonder
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which theorem do you mean specifically?
  $endgroup$
  – forgottenarrow
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was referring to the inequality; I thought I saw an interchange of $sum$ and $int,$ but now that I look closely, I see that I must have been dreaming, or something else was the case. You've simply added up infinitely many such inequalities -- nothing harmful in that at all. Whereupon, your well-earned upvote. It's a shame that OP got carried away by pictures, though.
  $endgroup$
  – Allawonder
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, that's a good thing to look out for and an easy mistake to make. Thanks!
  $endgroup$
  – forgottenarrow
  31 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

Consider the following summation:

$$
A=1cdot 1+frac12cdot1+frac13cdot1+frac14cdot1+...
$$

That's a sum of an infinite number of rectangles of height $frac1n$, where $ninmathbb{N}$, and width 1. Do you agree that's nothing more than the harmonic series? Moreover, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ is clearly less than $A$ and lies totally within $A$ (see the picture below). However, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ doesn't add up to a finite number. It goes to infinity. What should be happening with a piece of area that's even larger than that? Obviously, it must also be infinite. Therefore, the harmonic series diverges.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

$endgroup$

add a comment | 

3

$begingroup$

Consider the following summation:

$$
A=1cdot 1+frac12cdot1+frac13cdot1+frac14cdot1+...
$$

That's a sum of an infinite number of rectangles of height $frac1n$, where $ninmathbb{N}$, and width 1. Do you agree that's nothing more than the harmonic series? Moreover, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ is clearly less than $A$ and lies totally within $A$ (see the picture below). However, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ doesn't add up to a finite number. It goes to infinity. What should be happening with a piece of area that's even larger than that? Obviously, it must also be infinite. Therefore, the harmonic series diverges.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

$endgroup$

add a comment | 

$begingroup$

Consider the following summation:

$$
A=1cdot 1+frac12cdot1+frac13cdot1+frac14cdot1+...
$$

That's a sum of an infinite number of rectangles of height $frac1n$, where $ninmathbb{N}$, and width 1. Do you agree that's nothing more than the harmonic series? Moreover, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ is clearly less than $A$ and lies totally within $A$ (see the picture below). However, the area under the curve $f(x)=frac{1}{x}$ from $1$ to $infty$ doesn't add up to a finite number. It goes to infinity. What should be happening with a piece of area that's even larger than that? Obviously, it must also be infinite. Therefore, the harmonic series diverges.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

answered 4 hours ago

Michael RybkinMichael Rybkin

4,9441524

4

$begingroup$

It works using a simple inequality. If $f(x) leq M$ for $x in [a,b]$, then $int_a^b f(x),dx leq M(b-a)$. This is fairly easy to prove:

$$int_a^b f(x),dx leq int_a^b M,dx = M(b-a).$$

Notice that $1/x < 1/n$ on the interval $[n,n+1]$. Then,

$$infty = int_1^infty frac{1}{x}dx = sum_{k=1}^infty int_k^{k+1} frac{1}{x},dx leq sum_{k=1}^infty int_k^{k+1}frac{1}{k},dx = sum_{k=1}^infty frac{1}{k}.$$

share|cite|improve this answer

answered 4 hours ago

forgottenarrowforgottenarrow

33617

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The theorem as you state it is valid only for finite sums. However, you use it for an infinite sum in your explanation. As I'm sure you're aware, this doesn't always follow.
  $endgroup$
  – Allawonder
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which theorem do you mean specifically?
  $endgroup$
  – forgottenarrow
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was referring to the inequality; I thought I saw an interchange of $sum$ and $int,$ but now that I look closely, I see that I must have been dreaming, or something else was the case. You've simply added up infinitely many such inequalities -- nothing harmful in that at all. Whereupon, your well-earned upvote. It's a shame that OP got carried away by pictures, though.
  $endgroup$
  – Allawonder
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, that's a good thing to look out for and an easy mistake to make. Thanks!
  $endgroup$
  – forgottenarrow
  31 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

It works using a simple inequality. If $f(x) leq M$ for $x in [a,b]$, then $int_a^b f(x),dx leq M(b-a)$. This is fairly easy to prove:

$$int_a^b f(x),dx leq int_a^b M,dx = M(b-a).$$

Notice that $1/x < 1/n$ on the interval $[n,n+1]$. Then,

$$infty = int_1^infty frac{1}{x}dx = sum_{k=1}^infty int_k^{k+1} frac{1}{x},dx leq sum_{k=1}^infty int_k^{k+1}frac{1}{k},dx = sum_{k=1}^infty frac{1}{k}.$$

share|cite|improve this answer

answered 4 hours ago

forgottenarrowforgottenarrow

33617

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The theorem as you state it is valid only for finite sums. However, you use it for an infinite sum in your explanation. As I'm sure you're aware, this doesn't always follow.
  $endgroup$
  – Allawonder
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which theorem do you mean specifically?
  $endgroup$
  – forgottenarrow
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was referring to the inequality; I thought I saw an interchange of $sum$ and $int,$ but now that I look closely, I see that I must have been dreaming, or something else was the case. You've simply added up infinitely many such inequalities -- nothing harmful in that at all. Whereupon, your well-earned upvote. It's a shame that OP got carried away by pictures, though.
  $endgroup$
  – Allawonder
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, that's a good thing to look out for and an easy mistake to make. Thanks!
  $endgroup$
  – forgottenarrow
  31 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

It works using a simple inequality. If $f(x) leq M$ for $x in [a,b]$, then $int_a^b f(x),dx leq M(b-a)$. This is fairly easy to prove:

$$int_a^b f(x),dx leq int_a^b M,dx = M(b-a).$$

Notice that $1/x < 1/n$ on the interval $[n,n+1]$. Then,

$$infty = int_1^infty frac{1}{x}dx = sum_{k=1}^infty int_k^{k+1} frac{1}{x},dx leq sum_{k=1}^infty int_k^{k+1}frac{1}{k},dx = sum_{k=1}^infty frac{1}{k}.$$

share|cite|improve this answer

answered 4 hours ago

forgottenarrowforgottenarrow

33617

$endgroup$

share|cite|improve this answer

answered 4 hours ago

forgottenarrowforgottenarrow

33617

answered 4 hours ago

forgottenarrowforgottenarrow

33617

answered 4 hours ago

forgottenarrowforgottenarrow

33617