Ratio test — proof, limitations — Maths notes

WHAT it says

The quantity $|a_{n+1}/a_n|$ is the local shrink-ratio. $L$ is its long-run value.

WHY it works — derivation from scratch

The proof is comparison with a geometric series. We build it ourselves.

Case $L<1$ (convergence).

Why this step? Because $L<1$ , there's "room" between $L$ and $1$ . Pick a number $r$ in that gap.

Choose $r$ with $L < r < 1$ . By the definition of a limit, since $|a_{n+1}/a_n|\to L < r$ , eventually the ratio stays below $r$ : $\exists\,N \text{ such that } n\ge N \implies \left|\frac{a_{n+1}}{a_n}\right| < r.$

Why this step? This is just unpacking the limit: once we're past $N$ , every shrink-ratio is under $r$ .

Now chain the inequalities from $N$ onward: $|a_{N+1}| < r|a_N|,\quad |a_{N+2}| < r|a_{N+1}| < r^2|a_N|,\ \dots$ By induction: $|a_{N+k}| < r^k\,|a_N|.$

Why this step? Each term is at most $r$ times the one before, applied $k$ times gives $r^k$ .

So the tail is bounded term-by-term by a geometric series: $\sum_{k=0}^{\infty}|a_{N+k}| \;<\; |a_N|\sum_{k=0}^{\infty} r^k = \frac{|a_N|}{1-r} < \infty.$

Why this step? $\sum r^k$ converges because $0<r<1$ . A series whose terms are dominated by a convergent positive series converges (comparison test). Adding the finitely many terms before $N$ keeps it finite. Hence $\sum a_n$ converges absolutely. ∎

Case $L>1$ (divergence).

Choose $r$ with $1 < r < L$ . Then eventually $\left|\frac{a_{n+1}}{a_n}\right| > r > 1 \implies |a_{n+1}| > |a_n|.$

Why this step? Terms are growing past $N$ , so $|a_n|\not\to 0$ . But a necessary condition for any series to converge is $a_n\to 0$ (the term test). Since that fails, the series diverges. ∎

Figure — Ratio test — proof, limitations

WHY $L=1$ fails — the limitation

Other limitations:

Needs $a_n\neq 0$ (or eventually nonzero).
Tells you about absolute convergence; for conditional convergence (e.g. alternating harmonic) when $L=1$ , you need the alternating series test instead.
Best for terms with factorials or exponentials ( $n!$ , $r^n$ ), where ratios simplify cleanly.

Worked examples

Common mistake Steel-manned common errors

Mistake 1: " $L=1$ means the series diverges." Why it feels right: $L=1$ is the boundary where geometric domination just barely fails, so it "feels like" failure. Fix: $L=1$ means no information — could go either way ( $1/n$ vs $1/n^2$ ).

Mistake 2: Forgetting absolute values. Why it feels right: For positive series it doesn't matter. Fix: The proof uses $|a_{n+1}/a_n|$ ; for sign-changing series you must keep $|\cdot|$ , else $a_{n+1}/a_n$ can be negative and the geometric bound breaks.

Mistake 3: Using ratio test on $\sum 1/n^p$ . Why it feels right: It's a valid series, why not? Fix: It always gives $L=1$ — wasted effort. Recognize power-law terms and reach for the p-test.

Mistake 4: Concluding divergence just because $L=1$ and terms don't go to zero "fast enough." Fix: Use the actual term test ( $a_n\to 0$ ?) — that's a separate check.

Recall Feynman: explain to a 12-year-old

Imagine you're stacking blocks, and each new block is a fraction of the height of the one before. If every block is always less than, say, half the previous one, your tower stops growing — it reaches a finite height. The ratio test measures that fraction. If the fraction settles below 1, the tower is finite (converges). If it's above 1, blocks get bigger and the tower shoots to infinity (diverges). If the fraction sits exactly at 1, the blocks shrink so slowly we can't tell from the fraction alone whether the tower stops — we need a sharper ruler.

Flashcards

What is the ratio test limit

L

L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|

Ratio test verdict when

L<1

Converges absolutely.

Ratio test verdict when

L>1

L=\infty

Diverges.

Ratio test verdict when

L=1

Inconclusive — gives no information.

Which series does the ratio test compare against in its proof?

A geometric series

\sum r^k

In the proof for

L<1

, how is

r

chosen?

Any number with

L<r<1

, so eventually

|a_{n+1}/a_n|<r

Key inequality giving geometric domination?

|a_{N+k}|\le r^k|a_N|

for

n\ge N

Why does

L>1

force divergence?

Terms eventually grow, so

a_n\not\to 0

, violating the term test.

Two series both with

L=1

but opposite behaviour?

\sum 1/n

diverges;

\sum 1/n^2

converges.

What kinds of terms make the ratio test ideal?

Factorials (

n!

) and exponentials (

r^n

Ratio of

\sum 2^n/n!

and its

L

\frac{2}{n+1}\to 0

, converges.

Ratio of

\sum n!/10^n

and its

L

\frac{n+1}{10}\to\infty

, diverges.

What does the ratio test compute for a power series

\sum a_n x^n

Its radius of convergence (where

L<1

Why must absolute values be kept?

Sign-changing terms can give negative ratios, breaking the geometric bound; test concerns absolute convergence.

Connections

Geometric Series — the foundation the proof rests on.
Comparison Test — the mechanism that finishes the convergence case.
Term Test (nth-term divergence) — what kills the $L>1$ case.
Root Test — sibling test, often succeeds where ratio gives $L=1$ .
p-Series and Integral Test — the right tool when ratio test fails ( $L=1$ ).
Radius of Convergence — direct application to power series.
Raabe's Test — refinement for the $L=1$ borderline.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ratio test ka idea bilkul simple hai. Koi bhi infinite series $\sum a_n$ tab hi finite value pe converge karti hai jab uske terms kaafi tezi se chhote hote jaate hain. Ratio test poochta hai ek hi sawaal: "har agla term pichle term ka kitna guna hai?" Yeh ratio hai $|a_{n+1}/a_n|$ , aur iska long-run value ko hum $L$ kehte hain. Agar $L<1$ , matlab terms har baar ek fixed fraction se chhote ho rahe hain — yeh exactly geometric series jaisa behaviour hai, aur woh converge karti hai. Agar $L>1$ , terms badh rahe hain, toh series diverge kar jaati hai.

Proof ka dil yeh hai: agar $L<1$ , toh ek number $r$ chuno jo $L$ aur $1$ ke beech ho. Kuch index $N$ ke baad har ratio $r$ se chhota ho jaata hai, isliye $|a_{N+k}| \le r^k |a_N|$ . Yeh ek geometric series se "dominate" ho gaya, aur geometric series converge karti hai jab $r<1$ — comparison test se kaam ho gaya. $L>1$ wale case mein terms zero pe nahi jaate, aur koi bhi convergent series ke terms zero pe jaane chahiye, isliye diverge.

Sabse important baat — limitation: jab $L=1$ , test kuch nahi batata. Jaise $\sum 1/n$ (diverge) aur $\sum 1/n^2$ (converge) — dono ka $L=1$ hai! Isliye power-law (yaani $1/n^p$ type) series pe ratio test bekaar hai, wahan p-test ya integral test use karo. Ratio test asli kamaal factorials aur exponentials (jaise $n!$ , $2^n$ ) pe dikhata hai, kyunki wahan ratio bahut saaf simplify ho jaata hai. Yaad rakho: "Less than one — done, more than one — gone, equals one — no fun!"

Ratio test — proof, limitations

WHAT it says

WHY it works — derivation from scratch

WHY $L=1$ fails — the limitation

Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections