Root test: Worked examples — Maths deep dive

This is a child deep-dive of Root test. Here we do not re-derive the theorem; we hunt down every possible shape a root-test problem can take and work one example for each. If you meet a series in an exam, it falls into one of the rows below.

The scenario matrix

Throughout, let $a_{n}$ denote the $n$ -th term of the series $\sum a_{n}$ — the number sitting in the $n$ -th slot of the sum (for $\sum \frac{1}{n ^{2}}$ , the term $a_{n}$ is $\frac{1}{n ^{2}}$ ). Every root-test problem is decided by the single number $L = lim_{n \to \infty} n ∣ a_{n} ∣ .$ Here $n x$ ("the $n$ -th root of $x$ ") is the number that, multiplied by itself $n$ times, gives $x$ . It undoes raising to the power $n$ . That is the whole reason we use it: if a term secretly equals $r^{n}$ , then $n r^{n} = r$ pops the hidden ratio $r$ back out. We compute $L$ , then read the verdict: $L < 1$ converges, $L > 1$ diverges, $L = 1$ says nothing.

The cells below list every kind of input this machine can receive. Click the example label to jump to its full working.

Cell	Situation	What makes it distinct	Example
A	Clean $(stuff)^{n}$ , ratio $< 1$	root cancels power, $L < 1$	Ex 1
B	Exponential over polynomial, base $> 1$	$L > 1$ divergence	Ex 2
C	Polynomial over exponential	polynomial roots to $1$ , base $< 1$ survives	Ex 3
D	$L = 1$ degenerate (borderline)	test is silent, must switch tools	Ex 4
E	$L = 0$ super-fast decay	ratio effectively $0$ , converges hard	Ex 5
F	$L = \infty$ blow-up	terms explode, diverges	Ex 6
G	Negative / alternating terms	$	a_n
H	Word problem (bouncing / physical)	translate reality → series	Ex 8
I	Exam twist: $n$ in the exponent's exponent	$(1 + \frac{c}{n})^{n} \to e^{c}$	Ex 9

We will hit every one.

Cell A — clean power, ratio below 1

Worked example Ex 1 (Cell A) ·

$n = 1 \sum \infty (\frac{4 n - 1}{5 n + 2})^{n}$ Forecast: the whole bracket is raised to $n$ . Guess: does the fraction inside settle below, at, or above $1$ ? Jot down your guess before reading on.

Apply the root. $n ∣ a_{n} ∣ = n (\frac{4 n - 1}{5 n + 2})^{n} = \frac{4 n - 1}{5 n + 2} .$ Why this step? The term is a perfect $n$ -th power, so $n (\cdot)^{n} = (\cdot)$ — the root and the power annihilate. This is exactly the situation the root test is built for.
Take the limit. Divide top and bottom by $n$ : $\frac{4 - 1/ n}{5 + 2/ n} \to \frac{4}{5} .$ Why this step? As $n$ grows, $1/ n \to 0$ and $2/ n \to 0$ ; only the leading coefficients survive.
Read the verdict. $L = \frac{4}{5} < 1 \Rightarrow$ converges absolutely. Why this step? $L < 1$ means the terms eventually shrink faster than a convergent geometric series $\sum r^{n}$ with $\frac{4}{5} < r < 1$ .

Verify: plug $n = 100$ : $\frac{399}{502} = 0.7948 \approx 0.8$ . Trend confirms $L = 0.8 < 1$ . ✓

Cell B — exponential beats polynomial (divergence)

Cell C — polynomial over exponential (base below 1 survives)

Cell D — the silent case $L = 1$

Worked example Ex 4 (Cell D) ·

$n = 1 \sum \infty \frac{1}{n ( ln n ) ^{2}}$ (start at $n = 2$ ) Forecast: no obvious $n$ -th power here. Guess whether the root test will even decide this.

Compute the root. $n \frac{1}{n ( ln n ) ^{2}} = \frac{1}{n ^{1/ n} ( ln n ) ^{2/ n}} .$ Why this step? Root each factor: $n n = n^{1/ n}$ and $n (ln n)^{2} = (ln n)^{2/ n}$ .
Both roots $\to 1$ . $n^{1/ n} \to 1$ ; and $(ln n)^{2/ n} = e^{\frac{2 l n ( l n n )}{n}} \to e^{0} = 1$ since $\frac{l n ( l n n )}{n} \to 0$ . Why this step? Anything that grows slower than an exponential in $n$ roots to $1$ . Both factors qualify.
Verdict. $L = 1 \Rightarrow$ inconclusive — the root test cannot decide. Why this step? $L = 1$ is the blind spot: it can't distinguish convergent from divergent tails.
Switch tools. The Comparison test with the integral $\int_{2}^{\infty} \frac{d x}{x ( l n x ) ^{2}} = [- \frac{1}{l n x}]_{2}^{\infty} = \frac{1}{l n 2}$ (finite) shows the series converges.

Verify: the antiderivative check: $\frac{d}{d x} (- \frac{1}{l n x}) = \frac{1}{x ( l n x ) ^{2}}$ . ✓ And $L = 1$ exactly at the boundary. ✓

Cell E — decay so fast that $L = 0$

Cell F — blow-up, $L = \infty$

Cell G — negative / alternating terms

Cell H — the word problem

Worked example Ex 8 (Cell H) · A bouncing signal

A radar pulse loses power each hop. On hop $n$ it carries energy $E_{n} = (\frac{3}{4})^{n} n^{2}$ joules. Does the total energy over all hops converge to a finite amount? Forecast: the $\frac{3}{4}$ shrinks it, but $n^{2}$ pushes back. Who wins?

Model as a series. Total energy $= \sum_{n = 1}^{\infty} E_{n} = \sum n^{2} (\frac{3}{4})^{n} .$ Why this step? "Total over all hops" is literally the infinite sum of the per-hop energies.
Root each factor. $n n^{2} (\frac{3}{4})^{n} = (n n)^{2} \cdot \frac{3}{4} \to 1 \cdot \frac{3}{4} = \frac{3}{4} .$ Why this step? Polynomial $n^{2}$ roots to $1$ (Ex 2 logic); the geometric base $\frac{3}{4}$ survives.
Verdict. $L = \frac{3}{4} < 1 \Rightarrow$ converges: the total energy is finite. The physical geometric decay $\frac{3}{4} < 1$ beats the polynomial growth.

Verify (units + numerics): $E_{n}$ in joules; sum of joules is joules — dimensionally sound. Terms: $E_{1} = 0.75, E_{10} \approx 5.6, E_{30} \approx 0.16, E_{50} \approx 0.001$ — peak then decay to $0$ , total finite. ✓

The figure below makes this competition visible: the cyan bars are the per-hop energies $E_{n}$ (they rise to an early peak, then die away), while the amber curve is the running total — watch it flatten to a finite plateau, which is exactly what $L = \frac{3}{4} < 1$ guarantees.

Cell I — exam twist exponent inside an exponent

Recall Test yourself on the matrix

Which cell: $\sum (n / (3 n + 1))^{n}$ ? ::: Cell A — clean power, $L = 1/3 < 1$ , converges. Which cell: $\sum 7^{n} / n^{10}$ ? ::: Cell B — exponential over polynomial, $L = 7 > 1$ , diverges. Which cell: $\sum 1/ n^{p}$ (any $p$ )? ::: Cell D — $L = 1$ , inconclusive; use p-series. Which cell: $\sum 1/ (2 n)^{n}$ ? ::: Cell E — $L = 0$ , converges hard. Does an alternating sign change $L$ ? ::: No — root test uses $∣ a_{n} ∣$ , so the sign is erased (Cell G).