# The sum of a geometric series

One may be curious to find a sum of a geometric series &sum*q*^{k}. Does any such exist? That is, is it possible to associate a number with the expression *q* + *q*^{2} + *q*^{3} + ...

The answer depends on *q*. For *q*| < 1,*q* it does not.

The only way to define such a sum is by appealing to the theory of *limits*. By definition,

*q*+

*q*

^{2}+

*q*

^{3}+ ... =

*lim*

_{n→∞}∑

_{n},

where ∑_{n} is the *partial sum* of all the terms from the first and up to the n^{th} which is *q*^{n}.

The latter can be easily evaluated. Since

_{n}= 1 +

*q*+

*q*

^{2}+

*q*

^{3}+ ... +

*q*

^{n},

*q* ∑_{n} is found by multiplying that term-by-term by *q*:

*q*∑

_{n}=

*q*+

*q*

^{2}+

*q*

^{3}+

*q*

^{4}+ ... +

*q*

^{n +1},

Note that the two sums share all the terms except for the first (1) that appears only in the first sum ∑_{n} and the term *q*^{n +1}*q* ∑_{n}.

_{n}-

*q*∑

_{n}= 1 -

*q*

^{n +1},

or

_{n}= (1 -

*q*

^{n +1}) / (1 -

*q*).

We wish to define ∑ as *lim*_{n→∞}∑_{n}, i.e.

*lim*

_{n→∞}∑

_{n}=

*lim*

_{n→∞}(1 -

*q*

^{n +1}) / (1 -

*q*).

The latter exists and equals 1 / (1 - *q*) iff |*q*| < 1, for then and only then *lim*_{n→∞}*q*^{n +1} exists and when it does it equals 0.

For a more general series that starts with an arbitrary term *a*,

*a*+

*a*+

*q**a*

*q*^{2}+

*a*

*q*^{3}+ ... =

*a*/ (1 -

*).*

*q*|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

68354664