The geometric series ∑ xⁿ from n=0 to ∞ has different behaviors depending on the value of x. For -1 < x < 1, the series converges, and the sum is given by 1 / (1 - x). When x = 1, the series 1 + 1 + 1 + ... diverges, and there is no finite sum. When x = -1, the series 1 - 1 + 1 - 1 + ... also diverges, and the sum does not converge to a finite value. These special cases illustrate the importance of the condition -1 < x < 1 for the sum formula to be valid.
Understanding the Sum of \( x^n \): A Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number called the common ratio. In this case, the common ratio is \( x \), and the series can be expressed using sigma notation:
\( S = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \ldots \)
This series represents a pattern where you start with 1 and then multiply by \( x \) again and again, adding up all the terms.
Finding the Sum
If the common ratio \( x \) is between -1 and 1 (excluding -1 and 1), the terms get smaller and smaller, and the sum approaches a specific value. We can find this value using a clever trick:
1. Let \( S \) be the sum of the series: \( S = 1 + x + x^2 + x^3 + \ldots \)
2. Multiply both sides by \( x \): \( xS = x + x^2 + x^3 + x^4 + \ldots \)
3. Subtract the second equation from the first: \( S – xS = 1 \Rightarrow S = \frac{1}{1 – x} \)
This formula gives us the sum of the series when \( x \) is between -1 and 1. It’s a powerful insight into how the individual terms of the series combine to create a finite sum, and it has many applications in mathematics and real life.
Special Case: \( x = 1 \) in the Geometric Series
When \( x = 1 \), the geometric series becomes:
\( S = \sum_{n=0}^{\infty} 1^n = 1 + 1 + 1 + 1 + \ldots \)
Since each term is equal to 1, the series does not converge to a finite value. Instead, the sum grows without bound as more and more terms are added. In mathematical terms, we say that the sum diverges.
Expression \( \frac{1}{1 – x} \)
The expression \( \frac{1}{1 – x} \) gives the sum of the geometric series when \( x \) is between -1 and 1 (excluding -1 and 1). When \( x = 1 \), the expression becomes:
\( \frac{1}{1 – 1} = \frac{1}{0} \)
This expression is undefined because division by zero is not allowed in mathematics. It reflects the fact that the sum of the series diverges when \( x = 1 \), and there is no finite value that represents the sum.
In summary, when \( x = 1 \), both the geometric series and the expression \( \frac{1}{1 – x} \) indicate that the sum does not converge to a finite value. It’s a special case that illustrates the importance of the condition \( -1 < x < 1 \) for the sum formula to be valid.
Special Case: \( x = -1 \) in the Geometric Series
When \( x = -1 \), the geometric series becomes:
\( S = \sum_{n=0}^{\infty} (-1)^n = 1 – 1 + 1 – 1 + 1 – 1 + \ldots \)
The series alternates between 1 and -1, and there is no consistent pattern towards a specific value. In mathematical terms, we say that the sum does not converge to a finite value, and it is considered divergent.
Expression \( \frac{1}{1 – x} \)
The expression \( \frac{1}{1 – x} \) gives the sum of the geometric series when \( x \) is between -1 and 1 (excluding -1 and 1). When \( x = -1 \), the expression becomes:
\( \frac{1}{1 – (-1)} = \frac{1}{2} \)
However, this value does not represent the sum of the series when \( x = -1 \), as the series does not converge. The expression \( \frac{1}{1 – x} \) is only valid for \( -1 < x < 1 \), and it does not provide a meaningful result for \( x = -1 \).
In summary, when \( x = -1 \), both the geometric series and the expression \( \frac{1}{1 – x} \) indicate that the sum does not converge to a finite value. It’s another special case that illustrates the importance of the condition \( -1 < x < 1 \) for the sum formula to be valid.