. . . . . .

A "polynomial geometric" series is one in which the kth term is the product of an m-degree polynomial in k times x^k.

The way to find such a sum is to telescope the m-degree polynomial by taking successive differences.  That is, by multiplying the series by (1-x), the resulting series is a polynomial geometric series of degree m-1.  The first and last terms, don't cancel, however.  Then, repeating the process, i.e. multiplying again by (1-x), the series eventually becomes purely geometric, with m non-canceling terms at the beginning, and another m non-canceling terms at the end.  Then, multiply by (1-x) one more time, as you normally do with a geometric series, to make "most" of the terms cancel completely.

It turns out that the non-canceling terms become quite a challenge, especially if n < 2m -- that is, when the non-canceling terms overlap.  For this reason, I will confine the rest of this article to the case when n approaches infinity, so the non-canceling terms will occur only at the beginning of the series.  This doesn't actually cause any loss of generality, though, because a finite series can be expressed as the difference of two infinite series with different starting points.

Let S be the sum of an infinite series of this form,

Now, multiply the sequence by (1-x)^m+1, giving

S(1-x)^m+1 =

C(m+1,0)

∞ ∑ k=0

(a0 + a1k + a2k2 + ... + am-1km-1 + amkm) xk

- C(m+1,1)

∞ ∑ k=1

(a0 + a1(k-1) + a2(k-1)2 + ... + am-1(k-1)m-1 + am(k-1)m) xk

. . .

+ (-1)iC(m+1,i)

∞ ∑ k=i

(a0 + a1(k-i) + a2(k-i)2 + ... + am-1(k-i)m-1 + am(k-i)m) xk

. . .

+ (-1)m+1C(m+1,m+1)

∞ ∑ k=m

(a0 + a1(k-m) + a2(k-m)2 + ... + am-1(k-m)m-1 + am(k-m)m) xk

The beauty of this telescoping method is that all the terms of S(1-x)^m+1 except for the first m+1 terms cancel completely.  The only terms left are:

 . . . . . .

 . . . . . .

Related pages in this website

 . . . . . .

The webmaster and author of this Math Help site is Graeme McRae.