From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Newsgroups: sci.math
Subject: Re: Series total formulas
Date: 18 Sep 1998 12:01:37 -0400

In article <6tptql$8bn$1@news-2.news.gte.net>,
Jeffrey Sprankle <sprankle@gte.net> wrote:
>How does one go about finding the total formula for some series?
>For example,
>1+2+3+4+...+n = n(n+1)/2
>But how does one go about deriving this formula?

In addition to formula manipulation methods, there is a visual derivation
of this particular formula: just arrange the 1+2+...+n dots in a
triangular pattern, and attach to it a symmetric copy (it is the same as
the legendary nine-year Gauss procedure). I will demonstrate it on
1+2+3+4+5:

  *#####
  **####
  ***###
  ****##
  *****#

The rectangular arrangement has 5*6 objects, divided into equal numbers of
stars and sharps. So, the number of stars, which is 1+2+3+4+5, is also
half of 5*6, i.e. 5*(5+1)/2.  Similar picture can be made for n*(n+1)
objects.

In general, the problem of summing a finite series in a closed form, given
a formula for the general term ("symbolic summation"), is often harder
than "symbolic integration".

The easy (elementary) cases, such as polynomials, exponentials, and their
products, are treated well, among others, in the book

J.F. Steffensen: Interpolation, Chelsea Publ. Co. New York 1950, in
section 10.

Basically, one looks for a summation formula from the same family
(polynomial of degree one higher or so), with unknown coefficients, and
then matches those coefficients.

Some innocent-looking summation problems lead to "new" kinds of functions
(not contained in the supply of first-year Calculus examples): the
summation of the reciprocals (1/1+1/2+...+1/n) involves the Gamma function
and its derivative, for example.

Cheers, ZVK(Slavek).