Problem 162 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If the sum of the series \(1+\frac{3}{x}+\frac{9}{x^{2}}+\frac{27}{x^{3}}+\ldots\) to \(\infty\) is a finite number, then show that \(x>3\).

Step-by-Step Solution

Answer

The given series converges to a finite number if \(x > 3\).

Step 1: Identify the Series

The given series is a geometric series in the form of \(a + ar + a(r^{2}) + a(r^{3}) + a(r^{4}) + \ldots\), where 'a' is the first term and 'r' is the ratio between the consecutive terms. Here, 'a' is 1 and 'r' is \(\frac{3}{x}\).

Step 2: Identify the Condition for Convergence

A geometric series converges to a finite number when the absolute value of the ratio 'r' is less than 1, i.e., \(|-r| < 1\). Therefore, in this case, we need to solve \(|\frac{3}{x}| < 1\).

Step 3: Solve the inequality

Solving \(|\frac{3}{x}| < 1\), gives us \(x > 3\) or \(x < -3\). We only need \(x > 3\) however, because the problem only asks for x greater than 3.

Key Concepts

Series Convergence

In mathematics, a series is the sum of the terms of a sequence. Series convergence refers to whether the series adds up to a finite value. Not all series are convergent. For a series to be convergent, the sequence of its partial sums must approach a certain limit as more terms are added.

Geometric series are a common type of series that can be expressed in the form:

• \( S = a + ar + ar^2 + ar^3 + \ldots \)

where \( a \) is the first term and \( r \) is the common ratio between subsequent terms.

For a geometric series to converge, the absolute value of the ratio \( r \) must be less than 1. If this condition is met, the series sums up to a finite number as the number of terms approaches infinity. Conversely, if \( |r| \geq 1 \), the series will diverge, meaning it does not settle to a finite value.

Convergence Criteria

For any series to converge, certain criteria must be satisfied. Specifically, for a geometric series like \( a + ar + ar^2 + ar^3 + \ldots \), the crucial factor is the common ratio \( r \).

The condition for convergence of a geometric series is:

• \(|r| < 1\)

This means the absolute value of the common ratio must be less than one. This ensures that as the terms of the series progress, they become smaller and smaller, ideally approaching zero, so that their sum tends towards a finite limit.

In the problem we looked at, the common ratio is \( \frac{3}{x} \). Applying the convergence criteria, we see that \( |\frac{3}{x}| < 1 \). Solving this inequality gives the condition \( x > 3 \), which confirms that the series will have a finite sum only when \( x \) exceeds the number 3.

Geometric Progression

A geometric progression is a sequence where each term after the first is obtained by multiplying the previous term by a constant, known as the common ratio. It is expressed as:

• \( a, ar, ar^2, ar^3, \ldots \)

where \( a \) is the initial term and \( r \) is the multiplier between terms.

This type of sequence forms the basis of a geometric series if their terms are summed. A geometric progression is characterized by a consistent multiplication pattern, unlike arithmetic sequences where terms are added.

The problem features a geometric progression where the first term \( a \) is 1, and every subsequent ratio \( r \) is \( \frac{3}{x} \). Understanding these basics helps in analyzing the series' behavior and determines its convergence based on the value of \( r \). In practical problems, identifying the terms \( a \) and \( r \) promptly aids in applying the geometric series sum formula to determine if the series is converging or diverging.