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

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

- #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.

You don't have the parentheses in the right places. This should be 1 - 1/(n+1).Terms would cancel in that difference. The first series is a telescoping series.

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

- #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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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.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).

- #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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SBased 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.

- #11

Mark44

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This is incorrect - the left side is negative and the right side is close to 1 (hence positive).S_{n}- S_{n+1}= 1 - 1/(n+1).

- #12

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Are you sure about that?This is incorrect - the left side is negative and the right side is close to 1 (hence positive).

S

- #13

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If I write this in partial fractions I getHint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.

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?

The expression you have for SS_{n}- S_{n+1}= (1 + 1/2 + 1/3 + ... + 1/n) - (1/2 + 1/3 + ... + 1/n + 1(n+1)) = 1 - 1/(n+1).

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

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

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

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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Maybe.

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.

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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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.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}.

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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Your S_{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}

- #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.

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That's the sum I interpreted initially.

[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.

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

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

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