Power series — centre, radius of convergence, interval of convergence

WHAT is a power series?

WHY centre at $a$ ? Because convergence spreads symmetrically outward from one special point. That point is where convergence is guaranteed, so we measure distance $|x-a|$ from it.

WHAT is the radius / interval of convergence?

Three cases:

$R=0$ : converges only at $x=a$ .
$R=\infty$ : converges for all real $x$ .
$0<R<\infty$ : converges on a finite interval.

WHY is it always an interval? (Derivation from first principles)

This is the Cauchy–Hadamard idea. We test absolute convergence using the Root Test on the terms $b_n=|c_n(x-a)^n|=|c_n|\,|x-a|^n$ .

Formula Deriving

$R$ from the Root Test The Root Test (in its sharp form) looks at the limit superior $L=\limsup_{n\to\infty}\sqrt[n]{b_n}=\limsup_{n\to\infty}\sqrt[n]{|c_n|\,|x-a|^n}=|x-a|\cdot\underbrace{\limsup_{n\to\infty}\sqrt[n]{|c_n|}}_{\textstyle \beta}.$ Why this step? $\sqrt[n]{|x-a|^n}=|x-a|$ exactly, so the $x$ -part pulls out and only the coefficient part $\beta=\limsup\sqrt[n]{|c_n|}$ remains. Why $\limsup$ and not $\lim$ ? The ordinary $\lim$ may fail to exist (e.g. coefficients that oscillate), but $\limsup$ always exists in $[0,\infty]$ , so the formula works in full generality.

Root Test: converges if $L<1$ , diverges if $L>1$ . So $|x-a|\,\beta < 1 \iff |x-a| < \frac{1}{\beta}.$ Comparing with $|x-a|<R$ gives the Cauchy–Hadamard formula: $\boxed{\;\dfrac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{|c_n|}\;}$ (with the conventions $1/0=\infty$ and $1/\infty=0$ ). When the ordinary limit $\lim_{n\to\infty}\sqrt[n]{|c_n|}$ exists, it equals the $\limsup$ and you may use it.

Because the condition is $|x-a|<R$ — a distance condition — the solution set is automatically the symmetric interval around $a$ . That is why it is always an interval.

Formula Practical Ratio version

In practice the Ratio Test is easier (but it only delivers $R$ when the coefficient ratio settles down to a limit). Apply it to $b_n=|c_n||x-a|^n$ : $\lim_{n\to\infty}\frac{b_{n+1}}{b_n}=\lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}\,|x-a|=|x-a|\cdot\lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}.$ Converges when this $<1$ , so — provided the limit exists — $R=\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|.$ Why this step? Same logic: the $|x-a|$ factor separates out; setting the limit $<1$ gives the boundary $R$ . Caution (general form): if the ratio does not converge, the Ratio Test is silent; the guaranteed-to-work tool is then $R=1/\limsup_{n\to\infty}\sqrt[n]{|c_n|}$ . (Equivalently $R = 1/\limsup_{n\to\infty}|c_{n+1}/c_n|$ as an upper estimate.) The Root/ $\limsup$ form is the master formula; the Ratio limit is a convenient special case.

Figure — Power series — centre, radius of convergence, interval of convergence

HOW to find the interval (the recipe)

Apply Ratio Test to $|a_n|$ (with the $(x-a)^n$ included). Get $|x-a|\cdot k <1$ .
Solve for $|x-a|<R$ → this gives $R$ and the open interval $(a-R, a+R)$ .
Test each endpoint $x=a-R$ and $x=a+R$ by plugging in and using a series test (it becomes a number series).
Write the interval including/excluding endpoints accordingly.

Worked examples

Worked example Example 1 —

$\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$ Centre $a=2$ , coefficients $c_n=1/n$ . $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-2|^{n+1}/(n+1)}{|x-2|^n/n}=|x-2|\cdot\frac{n}{n+1}\to|x-2|.$ Why this step? The $|x-2|^n$ cancels leaving one factor; $n/(n+1)\to1$ . (Here the ratio limit exists, so the Ratio formula is valid.) Converge when $|x-2|<1\Rightarrow R=1$ , open interval $(1,3)$ . Endpoint $x=3$ : $\sum 1/n$ — harmonic, diverges. Endpoint $x=1$ : $\sum (-1)^n/n$ — alternating, converges (terms $\downarrow 0$ ). Interval: $\boxed{[1,\,3)}$ .

Worked example Example 4 —

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n (x+3)^n}{n^2\,4^n}$ Centre $a=-3$ . $\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x+3|^{n+1}/((n+1)^2 4^{n+1})}{|x+3|^n/(n^2 4^n)}=\frac{|x+3|}{4}\cdot\frac{n^2}{(n+1)^2}\to\frac{|x+3|}{4}.$ Converge when $|x+3|/4<1\Rightarrow |x+3|<4\Rightarrow R=4$ , interval $(-7,1)$ . Endpoint $x=1$ : $\sum (-1)^n/n^2$ — converges ( $p=2$ ). Endpoint $x=-7$ : $|x+3|=4$ gives $\sum (-1)^n(-4)^n/(n^2 4^n)=\sum 1/n^2$ — converges. Interval: $\boxed{[-7,\,1]}$ .

Forecast-then-Verify

Recall Forecast before computing

For $\displaystyle\sum \frac{(x-5)^n}{3^n}$ , predict $R$ before reading on. ▶ Coefficients $c_n=1/3^n$ , ratio $|c_n/c_{n+1}|=3^{n+1}/3^n=3$ , so $R=3$ , centred at $5$ . Geometric: both endpoints give $\sum(\pm1)^n$ which diverge. Interval $(2,8)$ . Did your forecast match?

What is the centre of

\sum c_n(x-a)^n

The point

a

; the series always converges there (gives

c_0

Cauchy–Hadamard formula for

1/R

(general, always valid)?

1/R=\limsup_{n\to\infty}\sqrt[n]{|c_n|}

, with

1/0=\infty

1/\infty=0

Ratio-test formula for

R

— when is it valid?

R=\lim_{n\to\infty}|c_n/c_{n+1}|

, ONLY when that limit exists; otherwise use the

\limsup

root formula.

What happens at

|x-a|=R

Inconclusive — must test each endpoint separately by substitution.

What does

R=0

mean?

Converges only at the centre

x=a

What does

R=\infty

mean?

Converges for all real

x

(e.g.

e^x

series).

Why is the convergence set always an interval?

The convergence condition is a distance condition

|x-a|<R

, symmetric about

a

Inside

|x-a|<R

, is convergence absolute or conditional?

Absolute convergence.

R

\sum n!\,x^n

0

(diverges for all

x\neq0

R

\sum x^n/n!

\infty

Why does Cauchy–Hadamard use

\limsup

instead of

\lim

Because

\limsup

always exists in

[0,\infty]

, so the formula gives

R

even when the ordinary limit fails.

Recall Feynman: explain to a 12-year-old

Imagine a campfire at point $a$ . The closer you stand, the warmer (the series "works"). There's a magic distance $R$ — inside it you're warm (converges), outside it you're freezing (diverges). Exactly on the circle of distance $R$ , it's a coin-flip: you have to walk to each edge and feel for yourself whether it's warm. The campfire is the centre, the magic distance is the radius, and the warm zone (edges maybe included) is the interval of convergence.

Connections

Ratio Test and Root Test — the engines that produce $R$ (Root/limsup is the master one).
Limit Superior and Limit Inferior — why $\limsup$ guarantees $R$ always exists.
Geometric Series — the prototype: $\sum x^n$ , $R=1$ .
Taylor and Maclaurin Series — power series built from a function's derivatives.
Absolute vs Conditional Convergence — inside the radius it's always absolute.
p-series and Alternating Series Test — the usual endpoint tools.
Term-by-term Differentiation and Integration — valid inside $(a-R,a+R)$ , $R$ unchanged.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Power series ko aise samjho jaise ek infinite-degree polynomial: $\sum c_n (x-a)^n$ . Yahan $a$ hai centre — yeh wo point hai jahan series hamesha converge karti hai (kyunki $x=a$ daalne par sirf $c_0$ bachta hai). Ab sawaal yeh hai: centre se kitni door tak series "kaam" karti hai? Us distance ko bolte hain radius of convergence $R$ .

$R$ nikalne ka asaan tareeka hai Ratio Test: $|a_{n+1}/a_n|$ ka limit lo, ismein $|x-a|$ bahar nikal aata hai. Formula: $R=\lim|c_n/c_{n+1}|$ — lekin yeh sirf tabhi valid hai jab yeh limit exist kare. Agar coefficients oscillate karein aur limit na aaye, to Ratio Test fail kar jaata hai. Tab hamesha kaam karne wala master formula hai Cauchy–Hadamard: $1/R=\limsup_{n\to\infty}|c_n|^{1/n}$ . $\limsup$ hamesha exist karta hai $[0,\infty]$ mein, isliye yeh formula har case mein $R$ de deta hai.

Sabse important baat — endpoints. Jab $|x-a|=R$ , tab Ratio Test ka limit exactly $1$ aata hai, jo inconclusive hota hai. Isliye dono endpoints $x=a-R$ aur $x=a+R$ ko alag-alag plug karke check karo (alternating series test, $p$ -series, ya divergence test se). Tabhi pata chalega ki interval mein bracket round hoga ( ya square [.

Yeh topic isliye zaroori hai kyunki Taylor/Maclaurin series, jaise $e^x$ , $\sin x$ , $\ln(1+x)$ — sab power series hain, aur unka interval of convergence batata hai ki approximation kahan tak valid hai. Bina radius jaane tum galat jagah series use kar baithoge!

Power series — centre, radius of convergence, interval of convergence

WHAT is a power series?

WHAT is the radius / interval of convergence?

WHY is it always an interval? (Derivation from first principles)

HOW to find the interval (the recipe)

Worked examples

Forecast-then-Verify

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections