What is the sum of a finite series with a constant p and variable n?

In summary, the conversation discusses the concept of "collapsing" or "telescoping series" to find the sum of a finite series that does not fall under the categories of arithmetic, geometric, or p-series. It is mentioned that for general values of p, there is no formula to solve such a series, and it becomes increasingly complicated as p increases. The conversation also mentions that for fractional values of p, there are no known formulas and the sum must be evaluated numerically.
  • #1
Luke77
42
0
Just wondering, is there a way to sort of "collapse" a finite series (to get the sum) that isn't classified as arithmetic, geometric or a p-series.
 
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  • #3
Since you use the word "collapse", there is always the "collapsing" or "telescoping series". For example, it is easy to show that the series [tex]\sum_{i=1}^n \frac{1}{n^2+ n}[/tex] sums to [itex]1- 1/(n+1)[/itex]. That is because, using "partial fractions", we can rewrite [tex]\frac{1}{n^2+ n}= \frac{1}{n}- \frac{1}{n+1}[/tex] so that each "1/k" term reappears as "-1/(k+1)" and cancels. The only terms that survive are the first, 1, and the last, -1/(n+1).
 
  • #4
Luke77 said:
Just wondering, is there a way to sort of "collapse" a finite series (to get the sum) that isn't classified as arithmetic, geometric or a p-series.

What kind of a series are you solving?
 
  • #5
I'm solving a series with just one coefficient, n, and an exponent. I'm aware that I can bring the coefficient "through" the integral and solve from there but I don't know what sort of formula to use.
I also know I can just write it out.
 
  • #6
HallsofIvy said:
Since you use the word "collapse", there is always the "collapsing" or "telescoping series". For example, it is easy to show that the series [tex]\sum_{i=1}^n \frac{1}{n^2+ n}[/tex] sums to [itex]1- 1/(n+1)[/itex]. That is because, using "partial fractions", we can rewrite [tex]\frac{1}{n^2+ n}= \frac{1}{n}- \frac{1}{n+1}[/tex] so that each "1/k" term reappears as "-1/(k+1)" and cancels. The only terms that survive are the first, 1, and the last, -1/(n+1).

I was hoping you would bring that up. For some reason, I really like telescoping series, but my series doesn't fall under it's description.
 
  • #7
Luke77 said:
I'm solving a series with just one coefficient, n, and an exponent. I'm aware that I can bring the coefficient "through" the integral and solve from there but I don't know what sort of formula to use.
I also know I can just write it out.

What, exactly, is the series? Is it [itex]S_p = \sum_{n=1}^N n^p,[/itex], or is it something else? It the series I wrote is the one you want, you are out of luck: for general p there is NO formula (although, of course, for integers p = 0,1,2,3, and a few more, you can derive results). However, the results get increasingly complicated as p increases. For example, when p = 3 we get
[tex] S_3 = \frac{1}{4} N^2 (n+1)^2,[/tex]
but for p = 10 we get
[tex] S_{10} = \frac{1}{66} N(N+1)(2N+1)(N^2-N-1)(3N^6+9N^5+2N^4 -11N^3 +3N^2 +10N-5),[/tex]
and it gets worse for larger p.

RGV
 
  • #8
Luke77 said:
I'm solving a series with just one coefficient, n, and an exponent. I'm aware that I can bring the coefficient "through" the integral and solve from there but I don't know what sort of formula to use.
I also know I can just write it out.

What, exactly, is the series? Is it [itex]S_p = \sum_{n=1}^N n^p,[/itex], or is it something else? It the series I wrote is the one you want, you are out of luck: for general p there is NO formula (although, of course, for integers p = 0,1,2,3---that are not really large---you can derive results). However, the results get increasingly complicated as p increases. For example, when p = 3 we get
[tex] S_3 = \frac{1}{4} N^2 (N+1)^2,[/tex]
but for p = 10 we get
[tex] S_{10} = \frac{1}{66} N(N+1)(2N+1)(N^2-N-1)(3N^6+9N^5+2N^4 -11N^3 +3N^2 +10N-5),[/tex]
and it gets worse for larger p. For fractional values of p, such as p = 1/2, etc., there are no known formulas. Basically, when we have such a series with known numerical values of N and p we just evaluate the sum numerically.

RGV
 
  • #9
Yes, it is the series you wrote: p stays the same, n is the term.
Thanks everyone.
 

1. What is the formula for finding the sum of a finite series?

The formula for finding the sum of a finite series is: S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term.

2. How do you find the sum of a finite series if the terms are not in a simple arithmetic pattern?

If the terms of a finite series are not in a simple arithmetic pattern, you can use the formula: S = (n/2)(a + a(r^n-1)), where r is the common ratio between terms.

3. Can you find the sum of an infinite series?

No, the sum of an infinite series cannot be found as it has no ending point. However, we can approximate the sum by calculating the sum of a large number of terms.

4. How is the sum of a finite series related to the area under a curve?

The sum of a finite series is related to the area under a curve in the sense that the sum can be represented graphically as the area under a curve. This is known as the integral of a function.

5. What is the significance of finding the sum of a finite series in mathematics?

The sum of a finite series is significant in mathematics as it helps us to understand the behavior of a sequence or a function. It also has many real-life applications, such as in financial calculations, physics, and computer science.

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