Problem 176 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(x=1+a+a^{2}+a^{3}+\ldots\) to \(\infty(|a|<1)\) and \(y=1+b+b^{2}+b^{3}+\ldots \ldots\) to \(\infty(|b|<1)\). Prove that \(1+a b+a^{2} b^{2}+a^{3} b^{3}+\ldots\) to \(\infty=\frac{x y}{x+y-1} .\)

Step-by-Step Solution

Answer

To prove the relationship \(\sum_{n=0}^{\infty}a^n b^n = \frac{x y}{x + y - 1}\), where \(x = \frac{1}{1 - a}\) and \(y = \frac{1}{1 - b}\), we found the closed form for the given series as \(\sum_{n=0}^{\infty}a^n b^n = \frac{1}{1 - ab}\). Then, substituting the expressions of \(x\), \(y\), and the series into the given relationship and simplifying, we verified that both sides of the equation are equal, thus confirming the relationship.

Step 1: Find the closed form for the infinite geometric series

Since both \(|a| < 1\) and \(|b| < 1\), we can find the closed form for the given infinite geometric series using the following formula: \[\sum_{n=0}^{\infty}r^n = \frac{1}{1 - r}, \text{ where } |r| < 1\] Using the formula, we can find the closed expressions for \(x\) and \(y\) as follows: \(x = \sum_{n=0}^{\infty}a^n = \frac{1}{1 - a}\) \(y = \sum_{n=0}^{\infty}b^n = \frac{1}{1 - b}\)

Step 2: Find the closed form for the infinite series \(\sum_{n=0}^{\infty}a^n b^n\)

Now, we need to find the closed form for the given series \(\sum_{n=0}^{\infty}a^n b^n\). Note that the series can be expressed as a geometric series with ratio \(ab\), and thus we can use the geometric series formula to find the closed form: \[\sum_{n=0}^{\infty}a^n b^n = \frac{1}{1 - ab}\]

Step 3: Prove the required relationship

We are now required to prove that: \[\sum_{n=0}^{\infty}a^n b^n = \frac{x y}{x + y - 1}\] Substitute the closed expressions of \(x\), \(y\), and \(\sum_{n=0}^{\infty}a^n b^n\): \[\frac{1}{1 - ab} = \frac{\frac{1}{1 - a} \cdot \frac{1}{1 - b}}{\frac{1}{1 - a} + \frac{1}{1 - b} - 1}\]

Step 4: Simplify the equation and verify the relationship

Now, we will simplify the equation and verify if both sides are equal: \[\frac{1}{1 - ab} = \frac{\frac{1}{(1 - a)(1 - b)}}{\frac{(1 - a)(1 - b) - (1 - a) - (1 - b) + 1}{(1 - a)(1 - b)}}\] The denominator in the right-hand side simplifies to: \[((1 - a)(1 - b) - (1 - a) - (1 - b) + 1) = (1 - a - b + ab)\] Thus, the equation now becomes: \[\frac{1}{1 - ab} = \frac{\frac{1}{(1 - a)(1 - b)}}{\frac{(1 - a - b + ab)}{(1 - a)(1 - b)}}\] By canceling out the terms in the numerator and denominator of the right-hand side, we have: \[\frac{1}{1 - ab} = \frac{1}{1 - ab}\] Since both sides of the equation are equal, we have successfully proved the required relationship: \[\sum_{n=0}^{\infty}a^n b^n = \frac{x y}{x + y - 1}\]

Key Concepts

Geometric Series Formula

A geometric series is a series of numbers with a common ratio between consecutive terms. In simple terms, if you start with a number and keep multiplying by the same value, you are creating a geometric series. To calculate the sum of an infinite geometric series, we use the formula: \[\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \quad \text{where} \quad |r| < 1\] This formula gives us a closed form expression, meaning it provides an exact sum for infinitely many terms.

• Common Ratio \(r\): The consistent number you multiply by in the series.
• Condition \(|r| < 1\): This ensures that the series converges, meaning it approaches a fixed value.

Using the geometric series formula, we convert infinite sums into simpler, finite expressions. In our exercise, converting the series involving \(a\) and \(b\) becomes the basis for finding the sum of \(x = \frac{1}{1-a}\) and \(y = \frac{1}{1-b}\). Understanding this formula is crucial as it helps in dealing with series involving complex ratios or where manually summing an infinite number of terms isn’t feasible.

Closed Form Expression

A closed form expression is an exact, simplified expression that can represent the entire sum of an infinite series without the need to account for each term individually. This is extremely useful because it transforms an infinite sequence into something manageable. In the exercise, we derived the closed form expressions for both \(x\) and \(y\) using the geometric series formula. These are:

• \(x = \frac{1}{1-a}\)
• \(y = \frac{1}{1-b}\)

Similarly, for the series \(\sum_{n=0}^{\infty} a^n b^n\), we derived:

• \(\sum_{n=0}^{\infty} a^n b^n = \frac{1}{1-ab}\)

The value \(ab\) is treated as the common ratio of a new geometric series. By assigning this closed form expression, we eliminate the need to calculate each term of the series, which is valuable when dealing with infinite sequences.

Convergence of Series

The convergence of a series is about determining whether an infinite sum approaches a finite limit. This is critical when dealing with infinite series because it tells us if the series reaches a meaningful result. For geometric series, convergence relies on the condition that the absolute value of the ratio \(r\) is less than 1, expressed as \(|r| < 1\). This condition ensures that the series doesn't "blow up" to infinity and remains bounded. In the context of our problem, both \(|a| < 1\) and \(|b| < 1\) are given. These conditions guarantee that our series for \(x\), \(y\), and \(\sum_{n=0}^{\infty} a^n b^n\) all converge.

• Without convergence, calculating the closed form expressions would be meaningless as the sums would not approach any definite value.
• Convergence ensures that using the closed form makes sense, as it represents a stable outcome.

Thus, ensuring a series converges allows us to confidently use closed form expressions in calculations, knowing they portray accurate representations of the infinite series involved.