Taming the Infinite: Differentiating Infinite Series | JEE Calculus

The Challenge of Infinite Differentiation

Differentiating infinite series is one of the most elegant yet challenging topics in calculus. While finite sums follow simple rules, infinite sums require careful consideration of convergence and uniform behavior.

⚠️ Common Misconception

"You can always differentiate an infinite series term-by-term" - This is FALSE and has led to many incorrect solutions in JEE Advanced.

🎯 JEE Advanced Significance

Appears in 3-8 mark problems in JEE Advanced
Essential for solving differential equations with series solutions
Used in Fourier analysis and power series methods
Tests conceptual depth rather than rote learning

1. Fundamental Concepts

What Makes Infinite Series Different?

Finite vs Infinite Sums

For finite sums: $\frac{d}{dx} \sum_{k=1}^n f_k(x) = \sum_{k=1}^n f_k'(x)$

For infinite sums: $\frac{d}{dx} \sum_{k=1}^\infty f_k(x) \neq \sum_{k=1}^\infty f_k'(x)$ in general

Counterexample

Consider $f(x) = \sum_{n=1}^\infty \frac{\sin(n^3x)}{n^2}$

The series converges for all $x$, but term-by-term differentiation gives:

$\sum_{n=1}^\infty n\cos(n^3x)$ which diverges for most $x$!

Pointwise vs Uniform Convergence

Pointwise Convergence

For each fixed $x$, the partial sums approach the limit.

$\forall x \forall \epsilon > 0 \exists N$ such that $|S_n(x) - S(x)| < \epsilon$ for $n > N$

Uniform Convergence

The convergence rate is independent of $x$.

$\forall \epsilon > 0 \exists N$ such that $|S_n(x) - S(x)| < \epsilon$ for all $x$ and $n > N$

💡 Key Insight

Uniform convergence is the key that unlocks term-by-term differentiation for infinite series.

2. Essential Theorems for Differentiation

JEE Advanced 2022 Must Know

Theorem 1: Term-by-Term Differentiation

Formal Statement

If $f_n(x)$ are differentiable on $[a,b]$ and:

$\sum f_n(x_0)$ converges for some $x_0 \in [a,b]$
$\sum f_n'(x)$ converges uniformly on $[a,b]$

Then $\sum f_n(x)$ converges uniformly to $f(x)$ and:

$$f'(x) = \sum_{n=1}^\infty f_n'(x)$$

Application Example

For $f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$ on $|x| \leq 1$:

Step 1: Check convergence at $x=1$: $\sum \frac{1}{n^2}$ converges

Step 2: Check derivative series: $\sum \frac{x^{n-1}}{n}$

Step 3: For $|x| \leq r < 1$, $\left|\frac{x^{n-1}}{n}\right| \leq \frac{r^{n-1}}{n}$

Step 4: By M-test, uniform convergence on $|x| \leq r$

Step 5: Therefore, $f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n}$ for $|x| < 1$

Theorem 2: Weierstrass M-Test

The Ultimate Convergence Tool

If $|f_n(x)| \leq M_n$ for all $x$ in domain, and $\sum M_n < \infty$, then $\sum f_n(x)$ converges absolutely and uniformly.

JEE Advanced Problem

Show that $f(x) = \sum_{n=1}^\infty \frac{\cos(nx)}{n^3}$ is differentiable for all $x$.

Step 1: Original series: $\left|\frac{\cos(nx)}{n^3}\right| \leq \frac{1}{n^3}$

Step 2: $\sum \frac{1}{n^3}$ converges ⇒ original series converges uniformly

Step 3: Derivative series: $\sum -\frac{\sin(nx)}{n^2}$

Step 4: $\left|\frac{\sin(nx)}{n^2}\right| \leq \frac{1}{n^2}$

Step 5: $\sum \frac{1}{n^2}$ converges ⇒ derivative series converges uniformly

Step 6: Therefore, term-by-term differentiation is valid

3. Power Series - The Special Case

The Radius of Convergence Magic

Power Series Theorem

For a power series $f(x) = \sum_{n=0}^\infty a_n x^n$ with radius of convergence $R > 0$:

$f(x)$ is differentiable on $(-R, R)$
$f'(x) = \sum_{n=1}^\infty n a_n x^{n-1}$ with the same radius $R$
This can be repeated infinitely many times!

Classic Example: Exponential Series

Consider $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ with $R = \infty$

Differentiation: $\frac{d}{dx}e^x = \sum_{n=1}^\infty \frac{n x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$

Index shift: Let $m = n-1$: $\sum_{m=0}^\infty \frac{x^m}{m!} = e^x$

Result: $\frac{d}{dx}e^x = e^x$ ✓

Advanced Power Series Techniques

JEE Advanced 2021 Problem

Find the sum of $\sum_{n=1}^\infty \frac{n^2 x^{n-1}}{2^n}$ for $|x| < 2$

Step 1: Recognize this as a derivative of a known series

Step 2: Start with $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $|x| < 1$

Step 3: Replace $x$ with $\frac{x}{2}$: $\sum_{n=0}^\infty \frac{x^n}{2^n} = \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x}$

Step 4: Differentiate once: $\sum_{n=1}^\infty \frac{n x^{n-1}}{2^n} = \frac{2}{(2-x)^2}$

Step 5: Differentiate again: $\sum_{n=1}^\infty \frac{n(n-1) x^{n-2}}{2^n} = \frac{4}{(2-x)^3}$

Step 6: Multiply by $x$ and manipulate to get $n^2$ term

Final Answer: $\frac{2x+4}{(2-x)^3}$

4. JEE Advanced Applications

JEE Advanced 2023 High Difficulty

Problem: Differential Equations with Series Solutions

Solve $y'' + xy = 0$ using power series

Step 1: Assume $y = \sum_{n=0}^\infty a_n x^n$

Step 2: Differentiate: $y' = \sum_{n=1}^\infty n a_n x^{n-1}$, $y'' = \sum_{n=2}^\infty n(n-1) a_n x^{n-2}$

Step 3: Substitute into DE:

$\sum_{n=2}^\infty n(n-1) a_n x^{n-2} + x \sum_{n=0}^\infty a_n x^n = 0$

Step 4: Adjust indices to match powers:

$\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n + \sum_{n=1}^\infty a_{n-1} x^n = 0$

Step 5: Solve recurrence relation

Step 6: Find general solution in series form

Fourier Series Differentiation

Differentiating Fourier Series

For $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty [a_n \cos(nx) + b_n \sin(nx)]$

⚠️ Warning: Fourier series can be differentiated term-by-term only if $f(x)$ is continuous and $f'(x)$ is piecewise smooth.

Valid case: $f'(x) = \sum_{n=1}^\infty [-n a_n \sin(nx) + n b_n \cos(nx)]$

When valid: When the differentiated series converges uniformly

🚨 Common Mistakes to Avoid

Conceptual Errors

Assuming term-by-term differentiation always works
Ignoring uniform convergence requirements
Differentiating at boundary points without checking
Confusing pointwise and uniform convergence

Calculation Errors

Incorrect index shifting in power series
Mishandling factorial expressions
Forgetting the radius of convergence changes
Errors in recurrence relation solutions

📚 Quick Reference Guide

When You CAN Differentiate Term-by-Term

Power series within radius of convergence
Series with uniform convergence of derivatives
Fourier series of smooth functions
Series satisfying Weierstrass M-test conditions

Essential Tests to Remember

Weierstrass M-test: For uniform convergence
Ratio Test: For power series radius
Abel's Test: For uniform convergence
Dirichlet's Test: For conditional convergence

🎯 Practice Problems

Test your understanding with these JEE-level problems:

1. Differentiate $f(x) = \sum_{n=1}^\infty \frac{x^n}{n(n+1)}$ and find its closed form.

Hint: Differentiate twice and recognize geometric series

2. Show that $\frac{d}{dx} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$

Hint: Recognize these as sine and cosine series

3. Find the interval where term-by-term differentiation is valid for $\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}$

Hint: Use Weierstrass M-test on the derivative series

Mastering Infinite Series Differentiation

You've learned the key concepts that separate JEE Advanced aspirants from the rest