- #1
Shackleford
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I'm not exactly sure what to do. S_{n} is the harmonic series. It diverges.
http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png
http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png
Shackleford said:I'm not exactly sure what to do. S_{n} is the harmonic series. It diverges.
http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png
Both of the sums shown in the linked-to image are finite sums.Shackleford said: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.
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.
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).
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 a finite sum.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.
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 S_{n} = 1 + 1/2 + 1/3 + ... + 1/n, which I say again, is a finite sum.
Shackleford said:S_{n} - S_{n+1} = 1 - 1/(n+1).
Mark44 said:This is incorrect - the left side is negative and the right side is close to 1 (hence positive).
LCKurtz said:Hint: Expand [itex]\frac 1 {n(n+1)}[/itex] into partial fractions.
Yes.Shackleford said:Are you sure about that?
Shackleford said: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).
?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.
Mark44 said: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.
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
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.
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.
How are you getting this?Shackleford said:Maybe.
S = 2S_{n} - S_{n+1}
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 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}.
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 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 to each term in the sum above to get a formula for S_{n}
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.
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.
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 + ...
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.
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.
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.
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.