Sum of an infinite geometric sequence (|r| < 1)

Let us first find the sum of following geometric sequence: 1 + 1/2 + 1/4 + 1/8 + 1/16 + .........

Let S = 1 + 1/2 + 1/4 + 1/8 + 1/16 + .........
then S - 1 = (1 + 1/2 + 1/4 + 1/8 + 1/16 + .........) - 1
which implies S - 1 = 1/2 + 1/4 + 1/8 + 1/16 + .........
So S - 1 = 1/2 times (1 + 1/2 + 1/4 + 1/8 + 1/16 + .........)
Since 1 + 1/2 + 1/4 + 1/8 + 1/16 + ......... = S
It follows that S - 1 = 1/2 * S
Therefore S - 1/2 * S = 1
So 1/2 * S = 1
Therefore S = 1 * 2 = 2

In general if we have a geometric sequence a, ar, ar^2, ar^3........ where a is the first term and r (common ratio) is such that |r| < 1

S = a + ar + ar^2 + ar^3........
S - a = ar + ar^2 + ar^3........
S - a = r * (a + ar + ar^2 + ar^3........)
S - a = r * S
S - r * S = a
S * (1 - r) = a
S = a/(1 - r) which denotes the sum of an infinite geometric sequence where |r| < 1

Example:
In the sequence 3 + 3/10 + 3/100 + ....
a = 3, r = 1/10
S = 3/(1 - 1/10) = 3/0.9 = 3.33333333... or 10/3


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