Term-by-term differentiation and integration of power series

1. What exactly are we claiming?

WHAT changes / WHAT stays?

The radius $R$ stays the same.
The behaviour at the endpoints $x=c\pm R$ can change (integration may gain convergence, differentiation may lose it). Always re-check endpoints separately.

2. WHY is the radius unchanged? (Derivation from scratch)

We never dump this — we earn it.

Derivation. The radius is governed (Cauchy–Hadamard) by $\frac{1}{R}=\limsup_{n\to\infty}|a_n|^{1/n}.$

For the differentiated series the coefficients are $b_n = (n+1)a_{n+1}$ (after re-indexing). Its radius $R'$ satisfies $\frac{1}{R'}=\limsup_{n\to\infty} |(n+1)a_{n+1}|^{1/n}.$

Now use the crucial limit, which we prove because it is the whole engine: $\lim_{n\to\infty} n^{1/n}=1.$

Therefore $|(n+1)|^{1/n}\to 1$ and the $\limsup$ value is unchanged: $\limsup |(n+1)a_{n+1}|^{1/n}=\limsup |a_{n+1}|^{1/n}=\frac1R.$ So $R'=R$ . The integration case is identical with factor $\frac{1}{n+1}$ , whose $1/n$ -th power also $\to 1$ . ∎

3. WHY can we swap $\frac{d}{dx}$ and $\sum$ at all?

4. Worked Examples

Worked example Example 2 — Integrate to get

$\arctan x$ Replace $x$ by $-x^2$ in the geometric series: $\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n},\quad |x|<1.$ Why? Because $\frac{1}{1-(-x^2)}$ matches the pattern; converges when $|-x^2|<1$ .

Integrate term by term from $0$ to $x$ : $\arctan x=\int_0^x\frac{dt}{1+t^2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots$ Why this step? $\int t^{2n}dt=\frac{t^{2n+1}}{2n+1}$ ; the constant is fixed by $\arctan 0=0$ . Radius still $1$ . Endpoint bonus: at $x=1$ the series $1-\frac13+\frac15-\cdots$ now converges (alternating), giving the Leibniz formula $\frac{\pi}{4}$ — convergence was gained at the endpoint by integrating.

Figure — Term-by-term differentiation and integration of power series

5. Forecast-then-Verify drill

Recall Forecast: differentiate

$\sum_{n=0}^{\infty}\frac{x^n}{n!}$ . What do you predict? Prediction: it should give itself ( $e^x$ ). Verify: $\frac{d}{dx}\sum\frac{x^n}{n!}=\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n!}=\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}=\sum_{m=0}^{\infty}\frac{x^m}{m!}.$ ✓ The series equals its own derivative — consistent with $e^x$ .

6. Flashcards

What happens to the radius of convergence when you differentiate a power series term by term?

It stays the same (

R

unchanged); only endpoint behaviour may change.

Why is

n^{1/n}\to 1

Because

\ln(n^{1/n})=\frac{\ln n}{n}\to 0

, so

n^{1/n}=e^{\ln n / n}\to e^0=1

What is the term-by-term derivative of

\sum a_n(x-c)^n

\sum_{n=1}^{\infty} n\,a_n (x-c)^{n-1}

What is the term-by-term integral of

\sum a_n(x-c)^n

C+\sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1}

Which can integration sometimes gain that differentiation can lose?

Convergence at an endpoint

x=c\pm R

What analytic property of power series justifies swapping

\frac{d}{dx}

and

\sum

Uniform convergence of the derivative series on compact subintervals of

(c-R,c+R)

Series for

\arctan x

derived how?

Integrate

\frac{1}{1+x^2}=\sum(-1)^n x^{2n}

term by term:

\sum \frac{(-1)^n}{2n+1}x^{2n+1}

Leibniz formula for

\pi/4

comes from?

Putting

x=1

in the

\arctan

series:

1-\frac13+\frac15-\cdots=\frac{\pi}{4}

Recall Feynman: explain to a 12-year-old

Imagine a super-long polynomial that never ends. Normal short polynomials are easy: to find the slope you just bring the power down, and to find area you bump the power up. The cool fact is that the never-ending one plays by the same simple rules as long as you stay in its "safe zone" (the inside of its interval). And the safe zone stays exactly the same size after you do this — only the two end-points (the edges) might start or stop behaving. So if you know one infinite-polynomial, you can make new ones for free by slope-ing or area-ing each piece.

Connections

Power series and radius of convergence
Cauchy–Hadamard theorem
Uniform convergence and the Weierstrass M-test
Taylor and Maclaurin series
Geometric series
Abel's theorem (endpoint behaviour)
Solving ODEs with power series

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, power series ko samajhne ka sabse easy tareeka hai: ise ek infinite polynomial maan lo. Aur polynomial differentiate/integrate karna toh bachche ka khel hai — power neeche le aao ya power upar badha do. Bilkul wahi rule yahan bhi chalta hai, har term pe alag-alag, jab tak tum interval of convergence ke andar ho. Sabse important baat: differentiate ya integrate karne ke baad bhi radius of convergence same rehta hai ( $R$ change nahi hota). Sirf do endpoints ka behaviour badal sakta hai — isliye endpoints hamesha dobara check karo.

Iska reason kya hai? Differentiate karne pe $n$ -th coefficient $n$ se multiply hota hai, integrate karne pe $(n+1)$ se divide. Par convergence ek geometric cheez hai jo $|a_n|^{1/n}$ pe depend karti hai, aur $n^{1/n} \to 1$ hota hai (kyunki $\frac{\ln n}{n}\to 0$ ). Matlab ye $n$ wala factor $n$ -th root mein gayab ho jata hai, isliye radius nahi badalta. Bas endpoint pe kabhi-kabhi convergence mil jaati hai (jaise integration ke baad) ya chhin jaati hai.

Practical fayda zabardast hai. $\frac{1}{1-x}=\sum x^n$ se shuru karo, $x$ ki jagah $-x^2$ daalo, phir integrate karo — seedha $\arctan x = x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots$ mil jata hai, bina koi mehnat ke! Aur $x=1$ rakho toh famous $\frac{\pi}{4}=1-\frac13+\frac15-\cdots$ (Leibniz). Isi tarah $\ln(1+x)$ bhi banta hai. Yaad rakhne ka mantra: DIRT — Differentiate/Integrate, Radius same, Test endpoints again. Bas yehi 80/20 funda hai is topic ka.

Term-by-term differentiation and integration of power series

1. What exactly are we claiming?

2. WHY is the radius unchanged? (Derivation from scratch)

3. WHY can we swap $\frac{d}{dx}$ and $\sum$ at all?

4. Worked Examples

5. Forecast-then-Verify drill

6. Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections