Problem 320 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a_{1}=1\) and \(a_{n+1}=\frac{4+3 a_{n}}{3+2 a_{n}}, n \geq 1\) and if \(\lim _{n \rightarrow \infty} a_{n}=a\), then find the value of \(a\).

Step-by-Step Solution

Answer

The value of the limit \(a\) as \(n\) approaches infinity of the sequence \(a_{n+1}=\frac{4+3 a_{n}}{3+2 a_{n}}\) is \(a=\sqrt{2}\).

Step 1: Define the Limit Value

Because the sequence is converging (i.e., getting closer and closer to a specific value as \(n\) approaches infinity), we can assume that \(a_{n+1}\) and \(a_{n}\) approach the same value, which we call \(a\).

Step 2: Substitute Limit into the Recursion Relation

Substitute \(a\) for \(a_{n+1}\) and \(a_{n}\) in the relation \(a_{n+1}=\frac{4+3 a_{n}}{3+2 a_{n}}\). This gives us the equation \(a=\frac{4+3 a}{3+2 a}\) for \(a\).

Step 3: Solve for \(a\)

Solving the equation for \(a\), firstly we can cross multiply to clear the fraction. This gives us \(a(3+2a)=4+3a\). Simplifying this equation leads to \(2a^{2}=4 \rightarrow a^{2}=2 \). Therefore, \(a=\sqrt{2} \) or \(a=-\sqrt{2}\).

Step 4: Validate the solution

However, based on the sequence definition \(a_{n+1}=\frac{4+3 a_{n}}{3+2 a_{n}}\), a negative solution is not valid, thus the value of \(a\) is \(a=\sqrt{2}\).

Key Concepts

Convergence of Sequences

When we talk about the convergence of sequences, we're exploring how a sequence behaves as it approaches a particular value indefinitely. A sequence is simply a list of numbers generated according to some rule. As the sequence progresses, if the terms get closer and closer to a specific number, we say that the sequence converges to that number. This concept is crucial because it helps us predict the long-term behavior of complex sequences.In our exercise, we have a sequence defined by a certain rule. As we let the term number, denoted by \( n \), grow very large, the sequence moves towards what we call a "limit." We denote this limit as \( a \). The key idea is that for large \( n \), both \( a_n \) and \( a_{n+1} \) are almost the same, meaning the sequence has converged to \( a \). Understanding this behavior forms the basis of finding the actual limit value when sequences are defined recursively or through complex relations.

Recursive Relations

Recursive relations in sequences allow us to generate terms based on previous ones. Think of it like a step-by-step instruction manual for building a sequence. In our problem, the sequence is defined recursively where each term depends on the prior term. The rule given in our exercise is \( a_{n+1}=\frac{4+3a_n}{3+2a_n} \). This formula elegantly encapsulates how you move from one term \( a_n \) to the next one \( a_{n+1} \).

• The starting point is given as \( a_1 = 1 \).
• Every subsequent term is computed using the recursive formula.

This recursive relationship simplifies the process of examining the sequence as it continues to infinity. It allows us to substitute a limit, knowing that at infinity, all the terms essentially collapse to a single value \( a \). This substitution leads to an equation that we can solve to find the value of \( a \). Recursive relations are like a bridge to understanding infinite behavior by examining finite steps.

Infinite Limits

Infinite limits reveal what happens to a sequence as it extends endlessly. For mathematical sequences, infinite limits provide valuable insights into their ultimate behavior. In our exercise, the concept of the infinite limit is key to determining the final value \( a \) that our sequence approaches. Consider that as \( n \) tends to infinity, instead of tracking each step, we concentrate on the end behavior - where exactly the sequence stabilizes. In mathematics, we express this stabilization or steady approach towards a value as \( \lim_{n \rightarrow \infty} a_n \). As sequences progress, infinite limits help establish whether they settle around a particular value (like \( \sqrt{2} \) in our problem) or diverge without settling.Recognizing infinite limits is essential in solving sequences defined with bounds or recursive methods. They offer a glimpse into the seemingly "boundless" nature of sequences but bring them back to a comprehensible, finite end.