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 Series

To 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)

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