Geometric Series Calculator
A geometric series is a sequence of the form
ra⁰, ra¹, ra², ra³,...
for some numbers r and a. In a geometric series, the ratio of consecutive terms is constant. Such sequences appear in discrete math problems such as compound interest, or processes with constant growth/decay rates. One of nice properties of geometric functions is that they are easy to sum. And if a is between -1 and 1, the infinite sum converges.
The summation formula is explained below, or you can use the calculator on the left. To use the calculator, enter the value of a as a fraction and the summation index values. For the infinite sum, enter "infinity" in the field for the upper index.
Formula for the Sum of a Geometric SeriesTo find the sum of a finite number of terms in a geometric series, consider the expression
G = a⁰ + a¹ + a² + ... + an
If we multiply both sides by a, we have
aG = a¹ + a² + a³ + ... + an + an+1
= (a⁰ + a¹ + a² + ... + an) - a⁰ + an+1
= G - 1 + an+1
Solving this expression for G gives us
G = (an+1 - 1)/(a - 1).
If |a| is less than 1 and n goes to infinity, the infinite sum is -1/(a-1), or 1/(1-a).
Example 1:Compute the sum
3 + 3(0.25) + 3(0.25)² + ... + 3(0.25)²⁰
First factor out the 3, since it can be multiplied at the end. This leaves us with
1 + 0.25 + 0.25² + ... + 0.25²⁰
= (0.25²¹ - 1)/(0.25 - 1)
= (1 - 0.25²¹)/0.75
= (4/3)[1 - (1/4)²¹]
Returning the factor of 3 gives us the final answer of
4[1 - (1/4)²¹]
Example 2:Let c = 1/π ≈ 0.31831. Simplify the infinite sum
B = c - c³ + c⁵ - c⁷ + c⁹ - ...
First, divide both sides by c:
B/c = 1 - c² + c⁴ - c⁶ + c⁸ - ...
This is equivalent to
B/c = (-c²)⁰ + (-c²)¹ + (-c²)² + (-c²)³ + (-c²)⁴ + ...
So we have
B/c = 1/[1 - (-c²)]
B/c = 1/[1 + c²]
B = c/[1 + c²]
B = (1/π)/[1 + 1/π²]
B = π/(1 + π²)
© Had2Know 2010