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_{n}is the harmonic series. It diverges.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png

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- Thread starter Shackleford
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http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png

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Hint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.

- #3

Mark44

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_{n}is the harmonic series. It diverges.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png

S

- #4

epenguin

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2 Actually

This should suggest something to you.

- #5

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Well, I realized that S_{n} - S_{n+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.

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.

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- #6

Mark44

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Both of the sums shown in the linked-to image are finite sums.Well, I realized that S_{n}- S_{n+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.

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

- #7

epenguin

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[tex]\frac{1}{n(n + 1)} = ? [/tex]

*Edit (Actually that is what Kurz is saying).*

Sooner or later you'll kick yourself.

Sooner or later you'll kick yourself.

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- #8

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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

Mark44

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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 SYeah, 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.

- #10

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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 S_{n}= 1 + 1/2 + 1/3 + ... + 1/n, which I say again, is afinitesum.

S

- #11

Mark44

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S_{n}- S_{n+1}= 1 - 1/(n+1).

This is incorrect - the left side is negative and the right side is close to 1 (hence positive).

- #12

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This is incorrect - the left side is negative and the right side is close to 1 (hence positive).

Are you sure about that?

S

- #13

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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

Mark44

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Yes.Are you sure about that?

S_{n}- S_{n+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 S

- #15

Mark44

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???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

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Yes.

The expression you have for S_{n+1}is wrong because it is missing a term. S_{n}is a sum of n terms, while S_{n+1}is a sum of n + 1 terms. S_{n}- S_{n+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

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???

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,

S

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- #18

Mark44

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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

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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 = 2S

- #20

Mark44

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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.

How are you getting this?Maybe.

S = 2S_{n}- S_{n+1}

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 S

[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 S

- #21

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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 S_{n}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 S_{n}= 1 + 1/2 + 1/3 + ... + 1/n. For this latter sum, they should have used a different letter altogether, maybe H_{n}.

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 [Broken]

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- #22

Mark44

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[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

- #23

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_{n}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 toeachtermin the sum above to get a formula for S_{n}

Your S

- #24

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[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

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[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.

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- #26

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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

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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))???

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