Monotonicity in Sequences and Series: A JEE Advanced Perspective
Master the crucial concept of monotonicity with convergence theorems, bounded sequences, and advanced problem-solving techniques for JEE Advanced.
Why Monotonicity is Crucial for JEE Advanced
Monotonic sequences and series form the backbone of convergence analysis in JEE Advanced. Understanding these concepts helps in:
- Proving convergence of complex sequences without explicit limits
- Solving inequality problems using monotonic behavior
- Analyzing recursive sequences common in JEE Advanced
- Understanding series convergence through sequence behavior
- Tackling 4-6 marks problems in every JEE Advanced paper
Definitions and Basic Properties
Monotonic Sequence Definition:
A sequence $\{a_n\}$ is called:
- Monotonically Increasing if $a_{n+1} \geq a_n$ for all $n$
- Monotonically Decreasing if $a_{n+1} \leq a_n$ for all $n$
- Strictly Increasing if $a_{n+1} > a_n$ for all $n$
- Strictly Decreasing if $a_{n+1} < a_n$ for all $n$
Testing Monotonicity:
Method 1: Check $a_{n+1} - a_n$
• If $a_{n+1} - a_n \geq 0$: Increasing
• If $a_{n+1} - a_n \leq 0$: Decreasing
Method 2: Check $\frac{a_{n+1}}{a_n}$ (for positive sequences)
• If $\frac{a_{n+1}}{a_n} \geq 1$: Increasing
• If $\frac{a_{n+1}}{a_n} \leq 1$: Decreasing
Example: Analyze $a_n = \frac{n}{n+1}$
Step 1: Compute $a_{n+1} - a_n = \frac{n+1}{n+2} - \frac{n}{n+1}$
Step 2: Simplify: $\frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} > 0$
Step 3: Since difference is always positive, sequence is strictly increasing
Bounded Monotonic Convergence Theorem
Theorem Statement:
Every bounded monotonic sequence converges.
- If $\{a_n\}$ is increasing and bounded above, it converges to its supremum
- If $\{a_n\}$ is decreasing and bounded below, it converges to its infimum
JEE Advanced Application:
This theorem is particularly useful for recursive sequences where explicit limit computation is difficult.
Example: $a_1 = 2$, $a_{n+1} = \frac{1}{2}(a_n + \frac{6}{a_n})$
Step 1: Show sequence is bounded below by $\sqrt{6}$ (by AM-GM inequality)
Step 2: Show $a_{n+1} \leq a_n$ (sequence is decreasing)
Step 3: By BMCT, sequence converges to $L$ where $L = \frac{1}{2}(L + \frac{6}{L})$
Step 4: Solve: $L^2 = 6 \Rightarrow L = \sqrt{6}$ (since $a_n > 0$)
Monotonic Series and Convergence Tests
Integral Test Connection:
For a positive, continuous, decreasing function $f(x)$ on $[1, \infty)$:
$\sum_{n=1}^{\infty} f(n)$ converges $\iff$ $\int_1^{\infty} f(x) dx$ converges
Example: Analyze $\sum_{n=1}^{\infty} \frac{1}{n^p}$
Step 1: Sequence $a_n = \frac{1}{n^p}$ is positive and decreasing for $p > 0$
Step 2: Apply integral test: $\int_1^{\infty} \frac{1}{x^p} dx$
Step 3: This integral converges if $p > 1$, diverges if $p \leq 1$
Step 4: Therefore, p-series converges for $p > 1$, diverges for $p \leq 1$
🚀 Advanced Problem-Solving Strategies
For Recursive Sequences:
- First prove boundedness (often using induction)
- Then prove monotonicity
- Use BMCT to guarantee convergence
- Find limit by solving $L = f(L)$
Common Pitfalls to Avoid:
- Assuming monotonicity without proof
- Forgetting to check boundedness
- Mixing up increasing/decreasing conditions
- Overlooking strict vs non-strict monotonicity
Concepts 4-5 Available in Full Version
Includes Advanced Recursive Sequences and Monotonicity in Functional Equations with JEE Advanced problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Determine if $a_n = \frac{n^2}{2^n}$ is monotonic and find its limit
2. Prove that $a_1 = 1$, $a_{n+1} = \sqrt{2 + a_n}$ converges and find its limit
3. Show that $\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$ converges using monotonicity arguments
Ready to Master Monotonic Sequences?
Get complete access to all concepts with step-by-step video solutions and JEE Advanced practice problems