Exploring the Divergence of the Harmonic Series with Ars Conjectandi

  • Thread starter Shackleford
  • Start date
  • Tags
    Series
In summary: That means that the polynomial on the left has to be identically equal to the one on the right.Grouping by powers of n gives1 = (A + B)n + AMore suggestively, this is 0n + 1 = (A + B)n + A
Physics news on Phys.org
  • #2
Hint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.
 
  • #4
1. But the difference between two divergent series is not necessarily divergent or infinite.

2 Actually write stuff out - write out Sn - Sn+1 as series and what does it equal? and write out the term typical term 1/n(n + 1) - how else would you naturally express that?

This should suggest something to you.
 
  • #5
Well, I realized that Sn - Sn+1 = S the first series. If I multiply that by 2 I get the sum identity inverse.

So, he's wanting an expression for the finite sum? I suspected it was finite, but I wasn't sure.

Terms would cancel in that difference. The first series is a telescoping series.

It simplifies to 1 - (1/n+1) for some n.
 
Last edited:
  • #6
Shackleford said:
Well, I realized that Sn - Sn+1 = S the first series. If I multiply that by 2 I get the sum identity inverse.

So, he's wanting an expression for the finite sum? I suspected it was finite, but I wasn't sure.
Both of the sums shown in the linked-to image are finite sums.
Shackleford said:
Terms would cancel in that difference. The first series is a telescoping series.

It simplifies to 1 - (1/n+1) for some n.

You don't have the parentheses in the right places. This should be 1 - 1/(n+1).
 
  • #7
[tex]\frac{1}{n(n + 1)} = ? [/tex]

Edit (Actually that is what Kurz is saying).

Sooner or later you'll kick yourself.
 
Last edited:
  • #8
Mark44 said:
Both of the sums shown in the linked-to image are finite sums.


You don't have the parentheses in the right places. This should be 1 - 1/(n+1).

Yeah, I have 1 - 1/(n+1) on my paper. If you take the limit as n goes to infinity it tends to 1. Since n does not go to infinity, then you have the expression for some finite n.
 
  • #9
Shackleford said:
Yeah, I have 1 - 1/(n+1) on my paper. If you take the limit as n goes to infinity it tends to 1. Since n does not go to infinity, then you have the expression for some finite n.
Based on the photo you uploaded, the question has nothing to do with limits. It could be that this is asked for in the problem itself, but it looks to me like what they're asking you to do is find Sn = 1 + 1/2 + 1/3 + ... + 1/n, which I say again, is a finite sum.
 
  • #10
Mark44 said:
Based on the photo you uploaded, the question has nothing to do with limits. It could be that this is asked for in the problem itself, but it looks to me like what they're asking you to do is find Sn = 1 + 1/2 + 1/3 + ... + 1/n, which I say again, is a finite sum.

Sn - Sn+1 = 1 - 1/(n+1).
 
  • #11
Shackleford said:
Sn - Sn+1 = 1 - 1/(n+1).

This is incorrect - the left side is negative and the right side is close to 1 (hence positive).
 
  • #12
Mark44 said:
This is incorrect - the left side is negative and the right side is close to 1 (hence positive).

Are you sure about that?

Sn - Sn+1 = (1 + 1/2 + 1/3 + ... + 1/n) - (1/2 + 1/3 + ... + 1/n + 1(n+1)) = 1 - 1/(n+1).
 
  • #13
LCKurtz said:
Hint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.

If I write this in partial fractions I get

1 = A(n+1) + Bn

If I continue, I don't get a nice expression.
 
  • #14
Shackleford said:
Are you sure about that?
Yes.
Shackleford said:
Sn - Sn+1 = (1 + 1/2 + 1/3 + ... + 1/n) - (1/2 + 1/3 + ... + 1/n + 1(n+1)) = 1 - 1/(n+1).

The expression you have for Sn+1 is wrong because it is missing a term. Sn is a sum of n terms, while Sn+1 is a sum of n + 1 terms. Sn - Sn+1 < 0.
 
  • #15
Shackleford said:
If I write this in partial fractions I get

1 = A(n+1) + Bn

If I continue, I don't get a nice expression.
?

The equation above is an identity that must be true for all n. That means that the polynomial on the left has to be identically equal to the one on the right.

Grouping by powers of n gives
1 = (A + B)n + A

More suggestively, this is
0n + 1 = (A + B)n + A
 
  • #16
Mark44 said:
Yes.


The expression you have for Sn+1 is wrong because it is missing a term. Sn is a sum of n terms, while Sn+1 is a sum of n + 1 terms. Sn - Sn+1 < 0.

I'm interpreting it differently. Why is it not the following?

sum of (1/k) from k = 1 to k = n

sum of (1/k+1) from k = 1 to k = n
 
  • #17
Mark44 said:
?

The equation above is an identity that must be true for all n. That means that the polynomial on the left has to be identically equal to the one on the right.

Grouping by powers of n gives
1 = (A + B)n + A

More suggestively, this is
0n + 1 = (A + B)n + A

I haven't worked partial fractions in a while. I reworked and got 1 = A and -1 = B.

According to what you're saying,

Sn - Sn+1 = (1 + 1/2 + 1/3 + ... + 1/n) - (1 + 1/2 + 1/3 + ... + 1/n + 1(n+1)) = -1/(n+1).
 
Last edited:
  • #18
Yes, that's better. Now, do you understand how all this ties into the real problem?
Namely, finding the sum
[tex]\frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

The whole business of partial fraction decomposition is intended to help you rewrite the individual terms in the sum above.
 
  • #19
Mark44 said:
Yes, that's better. Now, do you understand how all this ties into the real problem?
Namely, finding the sum
[tex]\frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

The whole business of partial fraction decomposition is intended to help you rewrite the individual terms in the sum above.

Maybe.

S = 2Sn - Sn+1
 
  • #20
Mark44 said:
Yes, that's better. Now, do you understand how all this ties into the real problem?
Namely, finding the sum
[tex]\frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

The whole business of partial fraction decomposition is intended to help you rewrite the individual terms in the sum above.

Shackleford said:
Maybe.

S = 2Sn - Sn+1
How are you getting this?

I worked this problem using the suggested hint and have something completely different for S.

One thing that bothers about the problem statement is their confusing use of S and Sn to represent unrelated things. For example, in the problem it is given that

[tex]S = \frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

Later on, they have Sn = 1 + 1/2 + 1/3 + ... + 1/n. For this latter sum, they should have used a different letter altogether, maybe Hn.
 
  • #21
Mark44 said:
How are you getting this?

I worked this problem using the suggested hint and have something completely different for S.

One thing that bothers about the problem statement is their confusing use of S and Sn to represent unrelated things. For example, in the problem it is given that

[tex]S = \frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

Later on, they have Sn = 1 + 1/2 + 1/3 + ... + 1/n. For this latter sum, they should have used a different letter altogether, maybe Hn.

Well, maybe that's my fault. I'm calling that S. The problem just gives the series. It doesn't call it S or anything.

Here's how I got that expression up there. You should be able enlarge it.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111105_142452.jpg
 
Last edited by a moderator:
  • #22
Then lets' not give names to 1 + 1/2 + ... + 1/n and 1 + 1/2 + ... + 1/n + 1/(n+1). And let's reserve Sn to mean the first n terms of the original sum. IOW,

[tex]S_n = \frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

As I see it, you aren't making the connection between the partial fractions business and the terms in this series. The partial fractions decomposition you did says that 1/(n(n+1)) = 1/n - 1/(n + 1).

Apply this formula to each term in the sum above to get a formula for Sn
 
  • #23
Mark44 said:
Then lets' not give names to 1 + 1/2 + ... + 1/n and 1 + 1/2 + ... + 1/n + 1/(n+1). And let's reserve Sn to mean the first n terms of the original sum. IOW,

[tex]S_n = \frac{1}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{3}{3 \cdot 4} + ... + \frac{n}{n \cdot (n + 1)}[/tex]

As I see it, you aren't making the connection between the partial fractions business and the terms in this series. The partial fractions decomposition you did says that 1/(n(n+1)) = 1/n - 1/(n + 1).

Apply this formula to each term in the sum above to get a formula for Sn

Your Sn is different than their first series. You can express your Sn as sum of n[(1/n) - 1/(n+1)] from 1 to n.
 
  • #24
This thread has gotten so bollixed up I can't tell whether Shackleford has solved the problem or not. The original sum was

[tex]\sum_{k=1}^n \frac 1 {k(k+1)} = \sum_{k=1}^n \left(\frac 1 {k}-\frac 1 {1+k}\right)[/tex]

Just write that out and cancel what you can and it's done.
 
  • #25
LCKurtz said:
This thread has gotten so bollixed up I can't tell whether Shackleford has solved the problem or not. The original sum was

[tex]\sum_{k=1}^n \frac 1 {k(k+1)} = \sum_{k=1}^n \left(\frac 1 {k}-\frac 1 {1+k}\right)[/tex]

Just write that out and cancel what you can and it's done.

That's the sum I interpreted initially.

You get 1 - 1/(n+1), right?

And using partial fractions you show that the above equals the desired series.
 
Last edited:
  • #26
Shackleford said:
That's the sum I interpreted initially.

You get 1 - 1/(n+1), right?

And using partial fractions you show that the above equals the desired series.

Yes. Partial fractions is how you express the single fraction as the difference of those two fractions.
 
  • #27
so, is Sn-Sn+1 supposed to equal 1/n - 1/(n+1)? So, that you then get Sn-Sn+1 = to 1/(n(n+1))?
 

Related to Exploring the Divergence of the Harmonic Series with Ars Conjectandi

1. What is the harmonic series?

The harmonic series is a mathematical series that is defined by the sum of the reciprocals of each positive integer. In other words, it is the sum of 1 + 1/2 + 1/3 + 1/4 + ...

2. What is the Ars Conjectandi?

The Ars Conjectandi, also known as The Art of Conjecturing, is a book written by mathematician Jacob Bernoulli in 1713. It discusses various theories and methods for making mathematical conjectures.

3. How does the harmonic series diverge?

The harmonic series diverges, meaning that the sum of its terms approaches infinity as the number of terms increases. This is because the reciprocals of the integers get smaller and smaller, but never reach zero.

4. What is the significance of exploring the divergence of the harmonic series?

Exploring the divergence of the harmonic series can provide insights into mathematical concepts such as infinite series, convergence, and divergence. It also has practical applications in fields such as physics and engineering.

5. What methods were used to explore the divergence of the harmonic series in Ars Conjectandi?

In Ars Conjectandi, Jacob Bernoulli used various mathematical techniques, including the method of exhaustion, to prove the divergence of the harmonic series. He also introduced the concept of the Bernoulli numbers, which are used in the study of infinite series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
244
  • Calculus and Beyond Homework Help
Replies
29
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
3K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Replies
2
Views
934
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
Back
Top