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# 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 expressionG =

*a⁰ + a¹ + a² + ... + a*

^{n}If we multiply both sides by

*a*, we have

*a*G =

*a¹ + a² + a³ + ... + a*

^{n}+ a^{n+1}= (

*a⁰ + a¹ + a² + ... + a*) -

^{n}*a⁰*+

*a*

^{n+1}= G - 1 +

*a*

^{n+1}Solving this expression for G gives us

G = (

*a*- 1)/(

^{n+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 sum3 + 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*= (-

*c*²)⁰ + (-

*c*²)¹ + (-

*c*²)² + (-

*c*²)³ + (-

*c*²)⁴ + ...

So we have

B/

*c*= 1/[1 - (-

*c*²)]

B/

*c*= 1/[1 +

*c*²]

B =

*c*/[1 +

*c*²]

B = (1/π)/[1 + 1/π²]

B = π/(1 + π²)

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