Ratio test — proof, limitations: Worked examples

The scenario matrix

Every series you can feed the ratio test falls into one of these cells. We will hit all of them.

Cell	What makes it special	$L$ outcome	Example #
A	Exponential ÷ factorial (clean cancellation)	$L = 0 < 1$ → converges	Ex 1
B	Factorial ÷ exponential (factorial wins)	$L = \infty > 1$ → diverges	Ex 2
C	Sign-changing / alternating terms (must use $∣ \cdot ∣$ )	$L < 1$ → abs. converges	Ex 3
D	Two exponentials, finite $0 < L < 1$	$L = r < 1$ → converges	Ex 4
E	Power-law $1/ n^{p}$ — the blind spot	$L = 1$ → inconclusive	Ex 5
F	Power series → radius of convergence, endpoints	$L = ∣ x ∣$ , then $L = 1$ boundary	Ex 6
G	Degenerate: a term is zero / ratio undefined	test doesn't apply — patch it	Ex 7
H	Real-world word problem (drug dosing)	$L = r < 1$ → finite build-up	Ex 8
I	Exam twist: $L = 1$ but Raabe rescues it	ratio fails, sharper tool needed	Ex 9
J	The limit $L$ doesn't exist (ratio oscillates)	no single $L$ → ratio test silent	Ex 10

Prerequisites we lean on: Geometric Series, Comparison Test, Term Test (nth-term divergence), Root Test, p-Series and Integral Test, Radius of Convergence, Raabe's Test.

Cell A — exponential over factorial

Worked example Example 1 ·

$n = 1 \sum \infty \frac{3 ^{n}}{n !}$ Forecast: factorials in the denominator grow faster than any exponential. Guess: converges, $L = 0$ .

Step 1. Write the ratio. $ρ_{n} = \frac{a _{n + 1}}{a _{n}} = \frac{3 ^{n + 1} / ( n + 1 )!}{3 ^{n} / n !} = \frac{3 ^{n + 1}}{3 ^{n}} \cdot \frac{n !}{( n + 1 )!} .$ Why this step? Dividing a fraction by a fraction is multiplying by the reciprocal; splitting into the power part and the factorial part lets each simplify on its own.

Step 2. Simplify each piece: $\frac{3 ^{n + 1}}{3 ^{n}} = 3$ and $\frac{n !}{( n + 1 )!} = \frac{1}{n + 1}$ (because $(n + 1)! = (n + 1) \cdot n!$ , the $n!$ cancels). $ρ_{n} = \frac{3}{n + 1} .$ Why this step? The advancing factorial contributes just a single new factor $(n + 1)$ , so the huge $n!$ cancels cleanly — that clean cancellation is exactly why the ratio test loves factorials.

Step 3. Take the limit. $L = lim_{n \to \infty} \frac{3}{n + 1} = 0 < 1 \Rightarrow converges absolutely.$ Why this step? A fixed numerator over a growing denominator collapses to $0$ .

Verify: This is the exponential series minus its $n = 0$ term: $\sum_{n = 1}^{\infty} 3^{n} / n! = e^{3} - 1 \approx 19.09$ — a finite number, consistent with convergence.

Cell B — factorial over exponential

Worked example Example 2 ·

$n = 1 \sum \infty \frac{n !}{5 ^{n}}$ Forecast: now the factorial is on top. Guess: diverges, $L = \infty$ .

Step 1. Ratio: $ρ_{n} = \frac{( n + 1 )!}{5 ^{n + 1}} \cdot \frac{5 ^{n}}{n !} = \frac{( n + 1 )!}{n !} \cdot \frac{5 ^{n}}{5 ^{n + 1}} = (n + 1) \cdot \frac{1}{5} .$ Why this step? Same split as before; here $\frac{( n + 1 )!}{n !} = n + 1$ grows, and $\frac{5 ^{n}}{5 ^{n + 1}} = \frac{1}{5}$ is a constant brake — but a constant can't stop a growing factor.

Step 2. Limit: $L = lim_{n \to \infty} \frac{n + 1}{5} = \infty > 1 \Rightarrow diverges.$ Why this step? $L = \infty$ falls in the divergence case; terms eventually blow up.

Verify: Argue growth from the ratio itself, not from one lucky term. For $n \geq 5$ we have $ρ_{n} = \frac{n + 1}{5} \geq \frac{6}{5} > 1$ , so every later term is strictly bigger than the one before: $a_{n + 1} > a_{n}$ for all $n \geq 5$ . A strictly increasing positive sequence cannot tend to $0$ , so $a_{n} \neq \to 0$ and the Term Test (nth-term divergence) forces divergence. (A single value like $a_{20} \approx 2.5 \times 1 0^{4}$ only illustrates this; the reason is the ratio staying above $1$ .) ✓

Cell C — sign-changing terms (absolute values matter)

Worked example Example 3 ·

$n = 1 \sum \infty \frac{( - 1 ) ^{n} 4 ^{n}}{( 2 n )!}$ Forecast: the $(- 1)^{n}$ makes signs flip, but the ratio test uses $∣ \cdot ∣$ which erases sign. The $(2 n)!$ crushes $4^{n}$ . Guess: absolutely converges, $L = 0$ .

Step 1. Absolute ratio (sign vanishes inside $∣ \cdot ∣$ ): $ρ_{n} = \frac{( - 1 ) ^{n + 1} 4 ^{n + 1} / ( 2 n + 2 )!}{( - 1 ) ^{n} 4 ^{n} / ( 2 n )!} = \frac{4 ^{n + 1}}{4 ^{n}} \cdot \frac{( 2 n )!}{( 2 n + 2 )!} .$ Why this step? $∣(- 1)^{n + 1} / (- 1)^{n} ∣ = 1$ , so keeping absolute values lets us treat the series like a positive one. Never drop the $∣ \cdot ∣$ on a sign-changing series: without it the raw ratio $a_{n + 1} / a_{n}$ would be negative, and a negative "ratio" cannot be compared to a geometric bound (which needs a positive $r < 1$ ). The whole geometric-domination argument silently breaks. That is why the definition insists on $∣ a_{n + 1} / a_{n} ∣$ .

Step 2. Simplify: $\frac{4 ^{n + 1}}{4 ^{n}} = 4$ , and $(2 n + 2)! = (2 n + 2) (2 n + 1) (2 n)!$ so $\frac{( 2 n )!}{( 2 n + 2 )!} = \frac{1}{( 2 n + 2 ) ( 2 n + 1 )}$ . $ρ_{n} = \frac{4}{( 2 n + 2 ) ( 2 n + 1 )} .$ Why this step? Expanding the factorial two steps down exposes exactly which factors cancel.

Step 3. Limit: $L = lim_{n \to \infty} \frac{4}{( 2 n + 2 ) ( 2 n + 1 )} = 0 < 1 \Rightarrow converges absolutely.$

Verify (two checks). (i) The signed sum is $cos (2) - 1$ ( $cos x = \sum (- 1)^{n} x^{2 n} / (2 n)!$ at $x = 2$ ): $cos 2 - 1 \approx - 1.416$ , finite → the series converges. (ii) The absolute sum $\sum \frac{4 ^{n}}{( 2 n )!} = cosh (2) - 1 \approx 2.762$ (since $cosh x = \sum x^{2 n} / (2 n)!$ at $x = 2$ ) is also finite — this is the check that actually confirms absolute convergence, exactly the claim the ratio test made. ✓

Cell D — two exponentials, finite ratio

Worked example Example 4 ·

$n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n} + n}$ (geometric figure) Forecast: for large $n$ the $+ n$ is negligible next to $4^{n}$ , so terms behave like $(3/4)^{n}$ . Guess: converges with $L = 3/4$ .

Step 1. Ratio: $ρ_{n} = \frac{3 ^{n + 1}}{4 ^{n + 1} + ( n + 1 )} \cdot \frac{4 ^{n} + n}{3 ^{n}} = 3 \cdot \frac{4 ^{n} + n}{4 ^{n + 1} + n + 1} .$ Why this step? Standard split; the messy denominators stay for now — we clean them by dividing.

Step 2. Divide top and bottom of the fraction by $4^{n}$ : $ρ_{n} = 3 \cdot \frac{1 + n / 4 ^{n}}{4 + ( n + 1 ) / 4 ^{n}} .$ Why this step? Dividing by the dominant term $4^{n}$ makes the sub-dominant $n / 4^{n} \to 0$ visible, isolating the true long-run ratio.

Step 3. Limit: $n / 4^{n} \to 0$ , so $L = 3 \cdot \frac{1 + 0}{4 + 0} = \frac{3}{4} < 1 \Rightarrow converges.$

Figure — Cell D (log-scale term plot). The horizontal axis is the term index

n = 1 \dots 12

; the vertical axis (logarithmic) is the term value

a_{n}

. The blue curve is the pure geometric sequence

(3/4)^{n}

; the orange curve is our actual terms

3^{n} / (4^{n} + n)

. The two curves lie almost on top of each other and both fall in a straight line on the log scale (the signature of geometric decay), with a thin green band marking the small gap where the geometric curve sits just above ours. Because our terms are dominated by a convergent Geometric Series of ratio

3/4 < 1

, the Comparison Test finishes the proof — matching the algebraic result

L = 3/4

Verify: compute $ρ_{n}$ numerically at $n = 15$ : it should sit very close to $0.75$ . ✓

Cell F — power series → radius of convergence & endpoints

Worked example Example 6 ·

$n = 1 \sum \infty \frac{x ^{n}}{n ^{2}}$ Forecast: the ratio test on a power series produces $L = ∣ x ∣$ , giving a Radius of Convergence. Then both endpoints $x = \pm 1$ hit $L = 1$ and need separate checks.

Step 1. Ratio (keep $∣ \cdot ∣$ since $x$ may be negative): $ρ_{n} = \frac{∣ x ∣ ^{n + 1}}{( n + 1 ) ^{2}} \cdot \frac{n ^{2}}{∣ x ∣ ^{n}} = ∣ x ∣ \cdot \frac{n ^{2}}{( n + 1 ) ^{2}} .$ Why this step? The $x$ -powers give $∣ x ∣$ ; the polynomial part is a $\to 1$ factor.

Step 2. Limit: $n^{2} / (n + 1)^{2} \to 1$ , so $L = ∣ x ∣$ . Converges for $∣ x ∣ < 1$ , diverges for $∣ x ∣ > 1$ → radius $R = 1$ . Why this step? $L < 1$ is exactly $∣ x ∣ < 1$ ; the boundary $∣ x ∣ = 1$ is left open.

Step 3 — endpoints ( $L = 1$ , test blind).

$x = 1$ : $\sum 1/ n^{2} = ζ (2) = π^{2} /6 \approx 1.645$ → converges (p-series, $p = 2$ ).
$x = - 1$ : $\sum (- 1)^{n} / n^{2}$ → absolutely convergent (its absolute version is the same $\sum 1/ n^{2}$ ) → converges. Why this step? At $L = 1$ you must abandon the ratio test and inspect each endpoint directly.

Interval of convergence: $[- 1, 1]$ .

Verify: at $x = 1$ the value is $π^{2} /6 \approx 1.645$ ; at $x = - 1$ the sum is $- π^{2} /12 \approx - 0.822$ — both finite. ✓

Cell G — degenerate input (a zero term)

Worked example Example 7 ·

$n = 1 \sum \infty a_{n}, a_{n} = \frac{1 - cos ( nπ )}{2} \cdot \frac{1}{2 ^{n}}$ Forecast: $1 - cos (nπ)$ is $0$ for even $n$ and $2$ for odd $n$ . So every even term is zero — the ratio $a_{n + 1} / a_{n}$ divides by $0$ and is undefined. The ratio test needs eventually-nonzero terms; here they never settle. We must patch it.

Step 1. Spot the zeros. $cos (nπ) = (- 1)^{n}$ , so $\frac{1 - ( - 1 ) ^{n}}{2}$ is $1$ (odd $n$ ) or $0$ (even $n$ ). Thus $a_{n} = {1/ 2^{n}, 0, n odd n even.$ Why this step? Before blindly forming a ratio, always check whether terms vanish — a zero denominator kills the formula.

Step 2. Keep only nonzero terms. The nonzero terms are exactly the odd indices, which we can list as $n = 1, 3, 5, \dots$ Introduce a new counter $k$ by setting $n = 2 k + 1$ ; as $k = 0, 1, 2, \dots$ this walks through every odd $n$ ( $k = 0 \to n = 1$ , $k = 1 \to n = 3$ , …). Then $2^{n} = 2^{2 k + 1}$ and the surviving series is $\sum_{k = 0}^{\infty} \frac{1}{2 ^{2 k + 1}} = \sum_{k = 0}^{\infty} \frac{1}{2} \cdot \frac{1}{4 ^{k}} .$ Why this step? Re-indexing to the nonzero terms (with the explicit substitution $n = 2 k + 1$ ) turns a degenerate case into a clean Geometric Series.

Step 3. This is geometric with first term $1/2$ and ratio $1/4 < 1$ → converges to $\frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{2}{3} .$ Why this step? Geometric sum $\frac{a}{1 - r}$ ; ratio $< 1$ guarantees convergence — no ratio test needed.

Verify: $\frac{2}{3} \approx 0.6667$ ; partial sum $1/2 + 1/8 + 1/32 + \dots$ climbs toward it. ✓

Takeaway: when terms hit zero, the ratio test's formula is undefined — strip zeros, or use the Comparison Test/Root Test (the root test tolerates zero terms because it uses $n ∣ a_{n} ∣$ , no division).

Cell H — real-world word problem (drug dosing)

Worked example Example 8 · Steady-state medicine

A patient takes a pill each day. The body clears a fraction so that only $70%$ of each dose remains after one day. If each pill delivers $200$ mg, how much drug is present in the body in the long run (just after the $n$ -th dose, as $n \to \infty$ )? Does the total build-up stay finite?

Forecast: doses stack but each old dose decays by $\times 0.7$ per day — a geometric pile-up. Ratio $0.7 < 1$ → finite.

Step 1. Model the amount from the dose taken $k$ days ago: it has decayed to $200 \cdot (0.7)^{k}$ mg. Just after today's dose, total is $A = k = 0 \sum \infty 200 (0.7)^{k}$ . Why this step? Each dose ages independently; summing over all past doses gives the current load.

Step 2. Ratio test the series $\sum 200 (0.7)^{k}$ : $ρ_{k} = \frac{200 ( 0.7 ) ^{k + 1}}{200 ( 0.7 ) ^{k}} = 0.7 = L < 1 \Rightarrow converges.$ Why this step? A pure geometric series has constant ratio equal to its common ratio — the cleanest possible ratio test.

Step 3. Sum it: $A = \frac{200}{1 - 0.7} = \frac{200}{0.3} \approx 666.7 mg (steady state).$ Why this step? Convergence (ratio $< 1$ ) lets us use the geometric closed form.

Verify: units are mg throughout; $200/0.3 = 666. \overline{6}$ mg — a bounded steady-state, so the drug does not accumulate to dangerous infinity. ✓

Cell I — exam twist: $L = 1$ , but Raabe rescues

Worked example Example 9 ·

$n = 1 \sum \infty \frac{1 \cdot 3 \cdot 5 \dots ( 2 n - 1 )}{2 \cdot 4 \cdot 6 \dots ( 2 n )} \cdot \frac{1}{n}$ Forecast: the double-product ratio simplifies to something $\to 1$ , so the ratio test will give $L = 1$ (inconclusive). This is the classic exam trap — you're expected to escalate to Raabe.

Step 1. Call the double product $c_{n} = \frac{1 \cdot 3 \dots ( 2 n - 1 )}{2 \cdot 4 \dots ( 2 n )}$ , so $a_{n} = c_{n} / n$ . Its ratio: $\frac{c _{n + 1}}{c _{n}} = \frac{2 n + 1}{2 n + 2}, \frac{a _{n + 1}}{a _{n}} = \frac{2 n + 1}{2 n + 2} \cdot \frac{n}{n + 1} .$ Why this step? Advancing the product by one step multiplies by the new numerator/denominator factor $(2 n + 1) / (2 n + 2)$ ; the extra $1/ n$ contributes $n / (n + 1)$ .

Step 2. Limit: $L = lim_{n \to \infty} \frac{2 n + 1}{2 n + 2} \cdot \frac{n}{n + 1} = 1 \cdot 1 = 1 \Rightarrow ratio test inconclusive.$ Why this step? Both factors approach $1$ — the crowded boundary strikes again.

Step 3 — escalate to Raabe. The terms are positive, so Raabe's hypothesis holds. Using the definition above, $\frac{a _{n}}{a _{n + 1}} = \frac{( 2 n + 2 ) ( n + 1 )}{( 2 n + 1 ) n} = \frac{2 n ^{2} + 4 n + 2}{2 n ^{2} + n},$ $\frac{a _{n}}{a _{n + 1}} - 1 = \frac{3 n + 2}{2 n ^{2} + n}, R = lim_{n \to \infty} n \cdot \frac{3 n + 2}{2 n ^{2} + n} = \frac{3}{2} .$ Since $R = \frac{3}{2} > 1$ → converges. Why this step? Raabe zooms into the rate at which the ratio approaches $1$ — the finer ruler the parent note promised when the ratio test is blind.

Verify: $R = 3/2 = 1.5 > 1$ , and the leading behaviour $a_{n} \sim \frac{C}{n ^{3/2}}$ (a $p = 3/2 > 1$ series) independently confirms convergence via p-Series and Integral Test. ✓

Cell J — when the limit $L$ doesn't even exist

Worked example Example 10 ·

$n = 1 \sum \infty a_{n}$ with $a_{n} = {1/ 2^{n}, 1/ 3^{n}, n even n odd$ Forecast: the ratios will alternate between two very different values, so no single $L$ exists — the ratio test goes silent and we finish another way.

Step 1. Look at consecutive ratios. Going from an odd $n$ to an even $n + 1$ and vice versa gives wildly different sizes, e.g. $\frac{a _{2}}{a _{1}} = \frac{1/4}{1/3} = \frac{3}{4}, \frac{a _{3}}{a _{2}} = \frac{1/27}{1/4} = \frac{4}{27} \approx 0.148,$ and this pattern of two alternating values repeats forever. Why this step? We compute a few $ρ_{n}$ to see that they do not approach a single number.

Step 2. Conclude the ratio test is inapplicable: $lim_{n \to \infty} ρ_{n}$ does not exist, so the $L < 1/ L > 1/ L = 1$ trichotomy has nothing to feed on. Why this step? No limit ⇒ no verdict; recognizing this saves you from writing a meaningless " $L$ ".

Step 3 — finish by comparison. Every term satisfies $0 \leq a_{n} \leq 1/ 2^{n}$ , and $\sum 1/ 2^{n}$ is a convergent Geometric Series. By the Comparison Test the series converges. Why this step? When the ratio test is silent, a direct geometric bound still settles it.

Verify: split by parity and sum two geometric series — even part $\sum_{m \geq 1} 1/ 4^{m} = 1/3$ , odd part $\sum_{m \geq 0} 1/ 3^{2 m + 1} = 3/8$ ; total $= 1/3 + 3/8 = 17/24 \approx 0.708$ , finite. ✓

Recall Which cell needs which tool?

Exponential-vs-factorial (Cells A–D) → ::: ratio test wins cleanly. Pure power law $1/ n^{p}$ (Cell E) → ::: p-series / integral test. Power series endpoints where $L = 1$ (Cell F) → ::: check each endpoint by hand. A term equals zero, ratio undefined (Cell G) → ::: strip the zeros / use the root test. Real-world geometric pile-up (Cell H) → ::: ratio equals the common ratio; use geometric sum. $L = 1$ from a factorial-product ratio (Cell I) → ::: Raabe's test. The ratio oscillates and $L$ doesn't exist (Cell J) → ::: ratio test is silent; use comparison (or limsup root test).

Connections

Ratio test — proof, limitations (index 4.3.12) — the parent: proof and theory.
Geometric Series — Cells D, G, H, J are geometric at heart.
Comparison Test — finishes Cells D, G and J.
Term Test (nth-term divergence) — the sanity check in Cell B.
p-Series and Integral Test — the correct tool for Cells E and F endpoints.
Radius of Convergence — what Cell F computes.
Root Test — the zero-tolerant sibling for Cell G and the limsup rescue for Cell J.
Raabe's Test — the sharper ruler for Cell I.

Worked examples — Ratio test — proof, limitations