Understanding the Infinite Series Problem

The question asks us to find the value of \(x\), which is defined by an infinite series:

\(x = 1 - y + y^2 - y^3 + \dots\) up to infinite terms.

We are also given the condition that \(|y| < 1\).

Identifying the Type of Series

Let's look closely at the terms in the series:

• The first term is \(1\).
• The second term is \(-y\).
• The third term is \(y^2\).
• The fourth term is \(-y^3\).
• And so on.

We can see a pattern here. Each term after the first is obtained by multiplying the previous term by \(-y\). This means the ratio between consecutive terms is constant and equal to \(-y\).

A series where the ratio between consecutive terms is constant is called a geometric series.

Analyzing the Geometric Series

For this specific geometric series:

• The first term, often denoted by \(a\), is \(1\).
• The common ratio, often denoted by \(r\), is \(-y\).

The series can be written as \(a + ar + ar^2 + ar^3 + \dots\), which matches \(1 + (1)(-y) + (1)(-y)^2 + (1)(-y)^3 + \dots = 1 - y + y^2 - y^3 + \dots\).

Condition for Convergence

An infinite geometric series converges (i.e., has a finite sum) if and only if the absolute value of the common ratio is less than 1 (\(|r| < 1\)).

In this problem, the common ratio is \(r = -y\). The given condition is \(|y| < 1\). Since \(|-y| = |y|\), the condition \(|y| < 1\) is equivalent to \(|-y| < 1\), or \(|r| < 1\). This confirms that the series converges and we can find its sum.

Formula for the Sum of an Infinite Geometric Series

The sum (S) of a convergent infinite geometric series with first term \(a\) and common ratio \(r\) (\(|r| < 1\)) is given by the formula:

\(S = \frac{a}{1 - r}\)

Calculating the Sum of the Series

Using the formula with \(a = 1\) and \(r = -y\), the sum \(x\) is:

\(x = \frac{1}{1 - (-y)}\)

Simplifying the denominator:

\(x = \frac{1}{1 + y}\)

Comparing with the Given Options

Now let's compare our derived value for \(x\) with the given options:

• Option 1: \(\(x = \frac{1}{{1 + y}}\)\)
• Option 2: \(\(x = \frac{1}{{1 - y}}\)\)
• Option 3: \(\(x = \frac{y}{{1 + y}}\)\)
• Option 4: \(\(x = \frac{y}{{1 - y}}\)\)

Our result, \(x = \frac{1}{1 + y}\), exactly matches Option 1.

Applying the values \(a=1\) and \(r=-y\) to the sum formula:

\(x = \frac{1}{1 - (-y)} = \frac{1}{1 + y}\)

Conclusion on the Value of x

Based on the analysis of the infinite geometric series and its sum formula, the value of \(x\) is \(\frac{1}{1 + y}\).

Revision Table: Key Concepts

Additional Information: Series and Convergence

Understanding different types of series is fundamental in mathematics. A series is essentially the sum of the terms of a sequence. Geometric series, like the one in this problem, are a specific type where the ratio between consecutive terms is constant. Other important types include arithmetic series and power series.

The concept of convergence is crucial for infinite series. If a series does not converge, it is said to diverge, meaning its sum does not approach a finite value. For a geometric series, the convergence depends solely on the common ratio \(r\). If \(|r| \ge 1\), the terms either grow larger or oscillate without settling, causing the sum to diverge.

The sum formula \(S = \frac{a}{1 - r}\) is a powerful tool, but it is only valid when the series converges. Always check the convergence condition \(|r| < 1\) before applying this formula for an infinite geometric series.