Ratio test — proof, limitations: Visual walkthrough

Step 1 — What is a series, as a picture?

WHAT. We turn each term $a_{n}$ into a coloured bar whose height is $∣ a_{n} ∣$ , then stack them.

WHY. "Does the series converge?" becomes a visible question: does the stacked tower reach a finite ceiling, or does it grow without bound? We use $∣ a_{n} ∣$ (the size, ignoring sign) because the vertical bars sizes are what pile up — signs are handled later.

WHAT IT LOOKS LIKE. In the figure, each bar is a term. If the bars shrink fast enough, the running total (the dashed line climbing on the right) flattens to a ceiling.

Figure — Ratio test — proof, limitations

Here $∣ a_{n} ∣$ = the height of one bar, and $S_{N}$ = total stacked height so far. The whole proof is about controlling how the bar heights shrink.

Step 2 — The one number we watch: the shrink-ratio

WHAT. We measure, at each step, how tall the next bar is compared to the current bar.

WHY division and not subtraction? Because a geometric shrink is multiplicative: "each bar is $70%$ of the one before" is a ratio statement, not a difference. A ratio of $0.7$ means "multiply by $0.7$ to get the next bar." That is exactly the language a geometric series speaks, and geometric series are the only ones we know how to sum in closed form. So division is the tool that matches the shape of the answer we're chasing.

WHAT IT LOOKS LIKE. In the figure, the ratio is the height of the next bar ÷ height of this bar — drawn as a bracket comparing two neighbouring bars.

If the ratio is small ( $0.3$ ): the next bar is a stub — fast shrink.
If the ratio is near $1$ : the next bar is almost as tall — barely shrinking.
If the ratio is above $1$ : the next bar is taller — the tower is growing.

Step 3 — The long-run ratio $L$

WHAT. We follow the shrink-ratio out to infinity and record the number it homes in on.

WHY. The first few bars can do anything — what decides convergence is the tail (the bars far out). $L$ throws away the noisy start and captures the tail's true shrink habit in one number.

WHAT IT LOOKS LIKE. Plot the shrink-ratio against $n$ . It wiggles early, then flattens toward a horizontal line at height $L$ . Three lines matter: $L$ below $1$ , $L = 1$ , $L$ above $1$ .

The whole verdict depends on where that flat line lands relative to the level $1$ .

Step 4 — $L < 1$ : pick a fence-rail $r$ in the gap

WHAT. We choose a number $r$ with $L < r < 1$ and draw it as a horizontal line above $L$ .

WHY a fixed $r$ , not just $L$ itself? Because "the ratio approaches $L$ " allows it to temporarily poke a hair above $L$ . But it can never poke above a rail that sits strictly higher. That strict headroom is what makes the next inequality permanent — we need a fixed wall, and $L < 1$ guarantees one exists (any $r$ in the gap works, e.g. the midpoint $r = \frac{L + 1}{2}$ ).

WHAT IT LOOKS LIKE. The settling curve of shrink-ratios dips under the rail $r$ and stays there forever after some point $n = N$ .

Step 5 — Chain the rails into a geometric fence

WHAT. Starting from bar $N$ , we cash in "next $<$ $r \times$ current" repeatedly.

WHY. One application gives $∣ a_{N + 1} ∣ < r ∣ a_{N} ∣$ . But we want a formula for the bar $k$ steps later, because summing a whole tail needs a pattern in $k$ , not just one step. Repeating the inequality $k$ times manufactures exactly that pattern: an $r^{k}$ .

WHAT IT LOOKS LIKE. The actual bars (violet) sit under the geometric fence-posts (orange) of heights $∣ a_{N} ∣, r ∣ a_{N} ∣, r^{2} ∣ a_{N} ∣, \dots$ — the fence droops geometrically and the bars never touch it.

Step 6 — Sum the fence: a finite ceiling

WHAT. We add up the whole geometric fence and get a closed number, then note our tower is shorter.

WHY this comparison? This is the Comparison Test mechanism: a stack of positive bars, each shorter than the matching bar of a convergent stack, must itself converge. Geometric is our "reference tower" because it's the one whose infinite height we can write down exactly.

WHAT IT LOOKS LIKE. Both running totals climb; the orange geometric total flattens to the ceiling $\frac{∣ a _{N} ∣}{1 - r}$ , and the violet real total stays below it — so it flattens too.

Step 7 — $L > 1$ : the tower grows, term test kills it

WHAT. We pick $1 < r < L$ and observe $∣ a_{n + 1} ∣ > r ∣ a_{n} ∣ > ∣ a_{n} ∣$ for large $n$ .

WHY the term test? There is an iron law: if a series converges, its terms must fall to $0$ . (If the bars didn't vanish, you'd keep adding non-vanishing amounts forever — infinite total.) Here the bars grow, so $∣ a_{n} ∣ \neq \to 0$ , so this necessary condition fails, so the series cannot converge — it diverges.

WHAT IT LOOKS LIKE. The bars climb upward, the shrink-ratio curve floats above the level $1$ , and the running total shoots off with no ceiling.

The case $L = \infty$ is the same story, even more violently: the ratio outgrows every rail.

Step 8 — $L = 1$ : the fence collapses (the honest limitation)

WHAT. We show two towers, both with shrink-ratio limit exactly $1$ , that end differently.

WHY show both? To prove $L = 1$ is not "secretly diverges" or "secretly converges" — it is truly ambiguous. The ratio test is blind to power-law terms $1/ n^{p}$ because their ratio $(\frac{n}{n + 1})^{p} \to 1$ always.

WHAT IT LOOKS LIKE. Two shrink-ratio curves both flatten onto the level $1$ from below — yet one tower reaches a ceiling ( $\sum 1/ n^{2}$ ) and the other never does ( $\sum 1/ n$ ).

The one-picture summary

Everything on one canvas: the number line of possible $L$ values, the fence you can (or cannot) build in each region, and the verdict.

$L < 1$ — a rail $r$ fits in the gap above $L$ ; bars droop under $r^{k} ∣ a_{N} ∣$ ; converges.
$L = 1$ — no rail fits either side; inconclusive.
$L > 1$ (incl. $\infty$ ) — bars grow, terms miss $0$ ; diverges.

Recall Feynman: the whole walkthrough in plain words

Picture stacking blocks. First I measure, for each block, what fraction of the previous block's height it has — that's the shrink-ratio. I follow that fraction far out the pile; the number it settles to is $L$ . If $L$ is under $1$ , there's spare room, so I nail a rail $r$ between $L$ and $1$ ; past some point every block is under $r$ times the one before, which forces the blocks under a geometric fence of heights $r^{k}$ times a starting block. That geometric fence has a known finite total, $\frac{∣ a _{N} ∣}{1 - r}$ , and since my real blocks are shorter, my tower has a ceiling — it converges. If $L$ is above $1$ , the blocks grow instead of shrink, so they never fall to zero, and a tower of non-vanishing blocks is infinitely tall — it diverges. If $L$ is exactly $1$ , no rail fits on either side: I've shown two real towers, $\sum 1/ n$ and $\sum 1/ n^{2}$ , both with $L = 1$ but one infinite and one finite — so the ratio alone honestly can't tell them apart. Less than one: done. More than one: gone. Equals one: no fun.

Connections

Geometric Series — the reference tower whose sum we borrow.
Comparison Test — the rule that lets a shorter tower inherit "finite".
Term Test (nth-term divergence) — what condemns the $L > 1$ case.
Root Test — sibling test using $n$ -th roots instead of ratios.
Radius of Convergence — where this whole machine is aimed for power series.
p-Series and Integral Test, Raabe's Test — the sharper rulers for the $L = 1$ blind spot.