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

[Luke77]

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.

 

[Mark44]

Not that I'm aware of.

 

[HallsofIvy]

Since you use the word "collapse", there is always the "collapsing" or "telescoping series". For example, it is easy to show that the series \sum_{i=1}^n \frac{1}{n^2+ n} sums to 1- 1/(n+1). That is because, using "partial fractions", we can rewrite \frac{1}{n^2+ n}= \frac{1}{n}- \frac{1}{n+1} 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).

 

[dimension10]

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?

 

[Luke77]

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.

 

[Luke77]

Since you use the word "collapse", there is always the "collapsing" or "telescoping series". For example, it is easy to show that the series \sum_{i=1}^n \frac{1}{n^2+ n} sums to 1- 1/(n+1). That is because, using "partial fractions", we can rewrite \frac{1}{n^2+ n}= \frac{1}{n}- \frac{1}{n+1} 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.

 

[Ray Vickson]

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 S_p = \sum_{n=1}^N n^p,, 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
S_3 = \frac{1}{4} N^2 (n+1)^2,
but for p = 10 we get
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),
and it gets worse for larger p.

RGV

 

[Ray Vickson]

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 S_p = \sum_{n=1}^N n^p,, 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
S_3 = \frac{1}{4} N^2 (N+1)^2,
but for p = 10 we get
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),
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

 

[Luke77]

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