Root test: Foundations — Maths deep dive

Before you can read one line of the parent note, you need every symbol it throws at you. Below is every piece of notation, in build-order. Nothing is used before it is drawn.

1. A series: what " $\sum a_{n}$ " even means

The full, precise notation is $\sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + a_{3} + \dots$ Read it piece by piece:

$\sum$ — the Greek capital sigma, meaning "add these all up".
$n = 1$ underneath — the index starts at position $1$ .
$\infty$ on top — it keeps going forever (no last term).
$a_{n}$ to the right — the general term, the rule giving the number at each position.

So $n = 1 \sum \infty a_{n}$ says exactly: "walk $n$ through $1, 2, 3, \dots$ endlessly, and add up every $a_{n}$ ." When the range is obvious we abbreviate it to just $\sum a_{n}$ .

$a_{n}$ = the general term = the number sitting at position $n$ . It is a rule, e.g. $a_{n} = \frac{1}{n ^{2}}$ means position 1 holds $\frac{1}{1}$ , position 2 holds $\frac{1}{4}$ , position 3 holds $\frac{1}{9}$ , …
$n$ = the index, a whole-number counter $1, 2, 3, \dots$ that walks through the positions.

Position n ::: the counting index 1, 2, 3, … that labels each term

\sum_{n = 1}^{\infty} a_{n}

, what does the

\infty

on top mean

the sum never ends — you add a term for every

n = 1, 2, 3, \dots

The symbol

\sum a_{n}

means

add up all the terms

a_{1} + a_{2} + a_{3} + \dots

forever

2. "Converges" and "diverges" — the two fates

In the figure above, the right panel's flattening curve is convergence; the climbing curve is divergence. The whole root test is a machine that decides which of these two fates a series meets — without you ever having to add it all up.

3. Absolute value $∣ a_{n} ∣$ and "absolute convergence"

Why does the root test use $∣ a_{n} ∣$ and not $a_{n}$ ? Because terms may be negative (e.g. $(- 1)^{n} / 2^{n}$ ), and the root test measures how big the terms are, not their direction. We strip the sign first.

The root's conclusion " $L < 1 \Rightarrow$ converges absolutely" is this exact idea: we tamed the sizes $∣ a_{n} ∣$ , so signs can't cause trouble.

4. The geometric series $\sum r^{n}$ — the yardstick everything is measured against

This is the first place the symbol $r$ appears — everything before spoke only of "a fixed fraction". The picture is a bouncing ball keeping a fixed fraction $r$ of its height each bounce.

Let's earn this instead of just stating it. We derive the exact value of the partial sum with the classic $S - r S$ trick.

Why this is the yardstick: geometric is the cleanest possible "shrinking" and we know its answer exactly. The root test's entire strategy is "make your series look like a geometric one, then read off $r$ ." See Geometric series for the full story.

Common ratio

r

\sum r^{n}

each term is

r

times the previous one

r^{0}

equals

1

(empty product convention)

The closed form of

1 + r + \dots + r^{N}

\frac{1 - r ^{N + 1}}{1 - r}

(for

r \neq = 1

)

\sum r^{n}

converges exactly when

∣ r ∣ < 1

, and then the total is

\frac{1}{1 - r}

5. Powers and the $n$ -th root $n \cdot$ — the star of the show

They are inverse operations, like a lock and its key: $n x^{n} = x (x \geq 0) .$

n y

answers the question

which number raised to the power

n

equals

y

n 2^{n}

equals

2

(the root cancels the

n

-th power)

6. Limits: the arrow $\to$ and $lim_{n \to \infty}$

Picture a dotted horizontal line at height $L$ ; the points $b_{n}$ hug that line ever tighter.

We need this because $n ∣ a_{n} ∣$ usually changes with $n$ — it's not one fixed ratio, it drifts. The limit $L$ is the settling value, the effective geometric ratio the term approaches.

b_{n} \to L

means

the terms

b_{n}

get and stay arbitrarily close to

L

n \to \infty

n n

tends to

1

7. $e$ and $ln$ — the machinery inside that limit

We only use one fact from them: they let us rewrite an awkward "variable exponent" like $n^{1/ n}$ into $e^{(l n n) / n}$ , turning a hard root into an easy exponent-limit. This is the exact trick behind $n n \to 1$ and behind the parent's $n 1/ n^{p} \to 1$ .

ln x

asks

e

raised to what power gives

x

Why rewrite

n^{1/ n}

e^{(l n n) / n}

it converts a variable-power root into an easy exponent limit

8. $sup$ and $lim sup$ — the "always-has-an-answer" limit

Picture a ceiling pressed down as low as it can go while still sitting on or above every point of $S$ .

sup S

the least upper bound — smallest number

\geq

every element of

S

T_{N} = sup {b_{n} : n \geq N}

N

grows

never rises (each tail is a subset), so it settles

lim sup

always exists because

the tail-ceilings

T_{N}

only decrease, so they always converge

9. Putting the symbol $L$ together — and proving the test

Now every piece of $L = n \to \infty lim sup n ∣ a_{n} ∣$ is earned:

$∣ a_{n} ∣$ — sizes of the terms, signs removed (§3),
$n$ — the un-do button for a geometric power, extracting the ratio (§5),
$lim sup / lim$ — the settling value of those extracted ratios (§6, §8).

So $L$ is the effective geometric ratio your series behaves like. Compare it to the yardstick $∣ r ∣ < 1$ (§4).

Formula Proof of the

$L < 1$ case (the bounding argument, in full) Suppose $L = lim sup n ∣ a_{n} ∣ < 1$ . Step 1 — pick a buffer $r$ . Since $L < 1$ there is room between $L$ and $1$ ; choose any fixed $r$ with $L < r < 1.$ Why: we need a geometric ratio that is (a) above the settling value so it eventually dominates the terms, and (b) below $1$ so its geometric series converges. A gap between $L$ and $1$ guarantees such an $r$ exists. Step 2 — the tail falls below $r$ . By the tail-ceiling picture of $lim sup$ (§8), the ceilings $T_{N} = sup {m ∣ a_{m} ∣ : m \geq N}$ decrease to $L < r$ . So for some cut-off $N$ , every term of the tail sits below $r$ : $n ∣ a_{n} ∣ < r for all n \geq N .$ Step 3 — raise to the $n$ -th power. Both sides are $\geq 0$ , so the inequality survives: $∣ a_{n} ∣ < r^{n} (n \geq N) .$ Step 4 — compare to the geometric yardstick. The tail $n \geq N \sum r^{n}$ is a convergent geometric series ( $0 \leq r < 1$ , §4). Since each $∣ a_{n} ∣$ is smaller, by the Comparison test the tail $\sum ∣ a_{n} ∣$ converges. Adding back the finite front chunk keeps it finite, so $\sum ∣ a_{n} ∣$ converges — i.e. $\sum a_{n}$ converges absolutely. $■$

Prerequisite map

Equipment checklist

Test yourself — reveal each only after you can say it out loud.

I can read every part of

\sum_{n = 1}^{\infty} a_{n}

sigma = add up;

n = 1

start index;

\infty

= forever;

a_{n}

= the term rule.

I can explain what

\sum a_{n}

means without the sigma symbol

It's the endless sum

a_{1} + a_{2} + a_{3} + \dots

of the terms given by the rule

a_{n}

I know the difference between converge and diverge

Converge = partial sums

S_{N}

settle on a finite number; diverge = they don't.

I know why we use

∣ a_{n} ∣

The test measures term size, ignoring sign, and gives absolute convergence.

I can derive the geometric partial sum

Multiply

S_{N}

r

, subtract:

(1 - r) S_{N} = 1 - r^{N + 1}

, so

S_{N} = \frac{1 - r ^{N + 1}}{1 - r}

I can state exactly when

\sum r^{n}

converges and its value

When

∣ r ∣ < 1

; then the total is

\frac{1}{1 - r}

(since

r^{N + 1} \to 0

I know why the geometric sum can start at

n = 0

n = 1

Adding/dropping the single term

r^{0} = 1

changes the total by a finite amount, never the convergence.

I can compute

n 2^{n}

and

n n^{2}

n 2^{n} = 2

;

n n^{2} = (n n)^{2} \to 1

I know why the

n

-th root is the right tool

It cancels an

n

-th power, so

n r^{n} = r

pulls the hidden ratio out.

I know the value of

lim n n

and why

1

, because

n^{1/ n} = e^{(l n n) / n}

and

\frac{l n n}{n} \to 0

I know what a supremum is and why

lim sup

always exists

Least upper bound; tail-ceilings

T_{N}

only decrease, so they always settle.

I can outline the

L < 1

proof

Pick

r

with

L < r < 1

; eventually

∣ a_{n} ∣ < r^{n}

; compare to convergent

\sum r^{n}

I can give two series with

L = 1

and opposite fates

\sum 1/ n

diverges,

\sum 1/ n^{2}

converges — both have

L = 1

I know what

L < 1

L > 1

L = 1

mean

Converge, diverge, inconclusive respectively.

Recall One-line recap

Every symbol in the root test exists to answer one question: does $a_{n}$ eventually behave like $r^{n}$ for some fixed $r < 1$ ? The $n$ -th root extracts that $r$ , the limit reads its settling value $L$ , and the geometric yardstick $∣ r ∣ < 1$ (proven via $S - r S$ ) decides the fate.

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