Root test

WHAT it says

We use $\limsup$ so the test always has a value, but for most exam problems the ordinary limit $\lim \sqrt[n]{|a_n|}$ exists and you use that.

WHY it works (derivation from scratch)

We build it entirely from the geometric series $\sum r^n$ , which converges iff $|r|<1$ .

Formula Case

$L<1$ — convergence by comparison Suppose $L=\lim\sqrt[n]{|a_n|} < 1$ . Pick any number $r$ strictly between $L$ and $1$ : $L < r < 1.$ Why this step? We need a fixed ratio $r$ that we can dominate the terms with, and that still makes a convergent geometric series — so $r$ must be below $1$ but above $L$ .

Since $\sqrt[n]{|a_n|} \to L < r$ , by definition of the limit there is an $N$ such that for all $n\ge N$ : $\sqrt[n]{|a_n|} < r.$ Why this step? Eventually the $n$ -th roots get squeezed below $r$ because $r$ is bigger than the limit $L$ .

Raise both sides to the $n$ -th power (both sides positive, so inequality is preserved): $|a_n| < r^n \quad (n\ge N).$ Now $\sum_{n\ge N} r^n$ is a convergent geometric series ( $0\le r<1$ ). By the comparison test the tail $\sum |a_n|$ converges, so $\sum a_n$ converges absolutely. $\blacksquare$

HOW to use it (worked examples)

Steel-manned mistakes

When to reach for it (80/20)

Root test limit formula

L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}

Conclusion when

L<1

Series converges absolutely.

Conclusion when

L>1

L=\infty

Series diverges (terms don't go to 0).

Conclusion when

L=1

Inconclusive — use another test.

Why does root test work for

L<1

Pick

r

with

L<r<1

; eventually

|a_n|<r^n

, and

\sum r^n

converges, so comparison gives convergence.

Why does root test work for

L>1

|a_n|>1

infinitely often, so

a_n\not\to0

; the

n

-th term test forces divergence.

What is

\lim_{n\to\infty}\sqrt[n]{n}

1

(since

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

When should you prefer the root test?

When the general term contains an

n

-th power

(\cdots)^n

or base

^n

For

\sum (2n+1)^n/(3n+5)^n

, find

L

L=2/3<1

, converges.

Why is

L=1

for every

\sum 1/n^p

\sqrt[n]{1/n^p}=e^{-p\ln n/n}\to 1

\ln n/n\to0

Recall Feynman: explain to a 12-year-old

Imagine each term in the sum is like the height you fall after each bounce of a ball. If every bounce keeps a fixed fraction $r$ of the previous height ( $r<1$ ), the total distance is finite — the ball stops. The root test is a clever trick to find that fraction: if a term looks like $r$ multiplied by itself $n$ times ( $r^n$ ), you "undo" the multiplying by taking the $n$ -th root, and out pops $r$ . If $r<1$ the bounces die out (sum is finite = converges). If $r>1$ the bounces grow forever (diverges). If $r$ comes out exactly $1$ , the trick can't tell — you need another tool.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, root test ka basic idea bohot simple hai. Hum jaante hain ki geometric series $\sum r^n$ tabhi converge karti hai jab $|r|<1$ . Ab kisi bhi series $\sum a_n$ ke baare mein hum poochhte hain: "Kya iske terms eventually $r^n$ jaise behave karte hain?" Agar haan, to us chhupe hue $r$ ko nikaalne ke liye hum $n$ -th root le lete hain — kyunki $\sqrt[n]{r^n}=r$ . Isliye $L=\lim \sqrt[n]{|a_n|}$ nikaalo: yeh limit hi effective ratio hai.

Rule yaad rakho: $L<1$ matlab terms tezi se chhote ho rahe hain, series converge. $L>1$ matlab terms bade ho rahe hain, $a_n$ zero tak nahi jaa raha, isliye diverge. Aur $L=1$ matlab test "Lost" — koi conclusion nahi, dusra test (p-series, comparison, integral) lagao. Proof bhi seedha hai: $L<1$ ho to ek $r$ choose karo jo $L$ aur $1$ ke beech ho, fir eventually $|a_n|<r^n$ , aur $\sum r^n$ converge karta hai — comparison se kaam khatam.

Root test kab lagana hai? Jab term mein $n$ -th power dikhe — jaise $(\text{kuch})^n$ ya $c^n$ . Tab $\sqrt[n]{\cdot}$ power ko cleanly cancel kar deta hai. Ek important trap: polynomial factors jaise $n^2$ ka $n$ -th root $n^{2/n}\to 1$ hota hai, $0$ nahi! Sirf exponential ka base bachta hai. Isliye $\sum n^2/2^n$ ke liye $L=1/2$ , converge karta hai. Bas yahi 20% concept 80% problems solve kar dega.

WHAT it says

WHY it works (derivation from scratch)

HOW to use it (worked examples)

Steel-manned mistakes

When to reach for it (80/20)

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections