Here's the standard proof: first look atfinitesums because isdefinedas the limit, as n goes to infinity of .

. Subtract a from both sides: .

Factor an "r" out of that sum: . Divide both sides by r: .

That isalmostwhat we started with- it is missing " " but we can add that to both sides:

.

Because these arefinitesums, all those are legal algebraic operations and we have now arrived at . Now solve for S.

That is the sum of afinitenumber, n, of terms. The only place n occurs in that formula is as " ". If r> 1 that will get larger and larger without bound as n increases so we cannot "take the limit" as n increases. If r< -1, then we have alternating positive and negative values but still getting larger and larger without bound in absolute value so we cannot "take the limit" as n increases.

If -1< r< 1, then r^n gets smaller and smaller- its limit is 0 and so the limit of S is .

Finally, if r= 1 we just have S= a+ a+ a+ ....+ a = na which gets larger and larger without bound (for all a except 0) and so has no limit. If r= -1, we get S= a- a+ a- a+/cdot/cdot/cdot a which is a if n is odd and 0 if n is even- still no limit.

That limit exist only if -1< r< 1, in which case the sum is or if a= 0 in which case the sum is 0 no matter what r is.