Integral test — proof, p-series: Visual walkthrough

A visual walkthrough of the parent result: the integral test. We build the whole proof out of rectangles and one falling curve — nothing else assumed. If you've never seen an integral sign, start at Step 1.

Step 1 — Turn the sum into a wall of rectangles

WHAT. Every term $a_{n}$ becomes a rectangle: width $1$ , height $a_{n} = f (n)$ . Line them up side by side. The total shaded area equals the sum $S_{N} = a_{1} + a_{2} + \dots + a_{N}$ .

WHY. Area is something we can see and bound. A raw sum of infinitely many numbers is invisible; a wall of rectangles is not. If we can trap the wall's area, we trap the sum.

PICTURE. Each bar sits on the number line from $x = n$ to $x = n + 1$ . Look at the height labels: the first bar is $f (1)$ tall, the next $f (2)$ , and so on — each shorter than the last, because our terms shrink.

Step 2 — Lay the smooth ramp over the bars

WHAT. Draw the curve $y = f (x)$ — a smooth ramp sliding downhill to the right. The area underneath it from $x = 1$ to $x = b$ is written $\int_{1}^{b} f (x) d x$ .

WHY. We can compute the ramp's area with calculus (that's what an integral does — it adds up infinitely thin slivers of area). We cannot compute the bar-sum directly. So we compare the two: whatever the ramp does, the bars are forced to copy.

PICTURE. The bars and the ramp share the same downhill shape. Some bars poke above the ramp; some hide below it. That poking-above / hiding-below is the whole game — Steps 3 and 4 pin it down exactly.

Step 3 — The tall bars: an OVER-estimate (upper sum)

WHAT. On the strip $[n, n + 1]$ , use the bar of height $f (n)$ — the value at the left edge. Since the ramp falls as we move right, $f (n)$ is the highest the ramp reaches on that strip. So this bar completely covers the ramp-area under that strip: $\int_{n}^{n + 1} f (x) d x \leq left-edge height f (n) .$

WHY. We want an upper bound on the ramp's area, built from our bars. Left endpoints on a decreasing curve always overshoot — that is exactly the inequality we need.

PICTURE. The bar is the big lavender box; the ramp-area is the smaller region tucked inside it. The thin sliver of bar above the curve is the "overshoot".

Now add these up for $n = 1, 2, \dots, N$ . The left sides join into one long integral, the right sides into the bar-sum: $ramp area, 1 to N + 1 \int_{1}^{N + 1} f (x) d x \leq = S_{N}, the bar wall n = 1 \sum N f (n) .$

Step 4 — The short bars: an UNDER-estimate (lower sum)

WHAT. Same strip $[n, n + 1]$ , but now use height $f (n + 1)$ — the value at the right edge. On a falling ramp that's the lowest height, so this short bar fits entirely under the ramp: $right-edge height f (n + 1) \leq \int_{n}^{n + 1} f (x) d x .$

WHY. We now want a lower bound on the ramp's area from our bars. Right endpoints on a decreasing curve always undershoot — the mirror image of Step 3.

PICTURE. The short coral bar is swallowed by the ramp-area; the sliver of ramp above the bar is the "gap".

Add these for $n = 1, 2, \dots, N - 1$ . On the left we get $f (2) + f (3) + \dots + f (N)$ , which is the whole sum minus its first bar: $all bars except the first S_{N} - f (1) \leq \int_{1}^{N} f (x) d x .$ Move $f (1)$ to the other side: $S_{N} \leq first bar f (1) + \int_{1}^{N} f (x) d x .$

Step 5 — Squeeze the wall between two ramps

WHAT. Stack Step 3 (a floor) under Step 4 (a ceiling). The bar-sum $S_{N}$ is now caught between two integrals: $floor \int_{1}^{N + 1} f \leq S_{N} \leq ceiling f (1) + \int_{1}^{N} f .$

WHY. This is the payoff. $S_{N}$ can't escape the floor or crash through the ceiling. Whatever fate the ramp's area meets — finite or infinite — the bar-sum is dragged to the same fate.

PICTURE. One number line, the wall of bars in the middle, the "floor" ramp tinted below and the "ceiling" ramp tinted above. The bars are wedged in the gap.

Step 6 — Read off the two verdicts

WHAT. Let $N$ run to infinity and watch the two edges.

Convergent case: if $\int_{1}^{\infty} f$ is a finite number $L$ , the ceiling never exceeds $f (1) + L$ . And $S_{N}$ only ever grows (we keep adding positive bars). A quantity that always rises but is capped from above must settle — so $\sum a_{n}$ converges.

Divergent case: if $\int_{1}^{\infty} f = \infty$ , the floor $\int_{1}^{N + 1} f$ climbs to infinity, and it's underneath $S_{N}$ , so it shoves $S_{N}$ up with it — $\sum a_{n}$ diverges. $■$

WHY. These are the only two things the ramp can do (rise forever, or level off). The squeeze converts each into the matching verdict for the sum.

PICTURE. Two panels: on the left, a levelling ramp with $S_{N}$ flattening onto a dashed cap; on the right, a rising floor dragging $S_{N}$ upward forever.

Step 7 — Feed in the p-series and split into cases

WHAT. Take $a_{n} = \frac{1}{n ^{p}}$ , so $f (x) = x^{- p}$ . It's positive, continuous, and decreasing for every $p > 0$ — hypotheses met. By Step 6, the sum's fate = the ramp's fate. Compute the ramp area.

Case $p \neq = 1$ : $\int_{1}^{\infty} x^{- p} d x = lim_{b \to \infty} \frac{b ^{1 - p} - 1}{1 - p} .$

$p > 1$ : exponent $1 - p < 0$ , so $b^{1 - p} \to 0$ . Area $= \frac{1}{p - 1}$ (finite) $\Rightarrow$ converges.
$0 < p < 1$ : exponent $1 - p > 0$ , so $b^{1 - p} \to \infty$ . Area infinite $\Rightarrow$ diverges.

Case $p = 1$ (the Harmonic series): the power rule breaks, the integral becomes the logarithm: $\int_{1}^{\infty} \frac{d x}{x} = lim_{b \to \infty} ln b = \infty \Rightarrow diverges .$

WHY. The sign of the exponent $1 - p$ decides whether the ramp's tail vanishes or explodes — that single sign is the razor's edge at $p = 1$ .

PICTURE. Three ramps overlaid: a steep one ( $p = 2$ , finite tail area, tinted and closed off), the borderline ( $p = 1$ , tail area creeping up forever), and a gentle one ( $p = \frac{1}{2}$ , tail flooding to infinity).

The one-picture summary

Everything in one frame: the wall of bars ( $S_{N}$ ), the ramp ( $y = f (x)$ ), the tall left-endpoint bars poking above (upper sum), the short right-endpoint bars hiding below (lower sum), and the squeeze arrows pinning $S_{N}$ between the two integrals.

Recall Feynman: retell the whole walkthrough in plain words

I want to know if adding $a_{1} + a_{2} + a_{3} + \dots$ forever lands somewhere or runs off to infinity. I can't add forever, so I draw each term as a block of width $1$ and height $a_{n}$ — now the sum is just the shaded area of a wall of blocks. Next to the blocks I lay a smooth slide $y = f (x)$ whose area I can compute with an integral. On this downhill slide, if I use the left edge of each strip the block is a bit too tall — it covers more than the slide (upper sum). If I use the right edge the block is a bit too short — the slide covers more than it (lower sum). So the block-wall is squeezed between "the slide's area" and "the slide's area plus one extra first block". Because the wall only ever grows (all blocks point up), if the slide's area is finite the wall is capped and must settle; if the slide's area is infinite the wall is shoved up to infinity too. Same fate, always. Then I try $1/ n^{p}$ : its slide is $x^{- p}$ , whose tail area shrinks to nothing only when $p > 1$ and floods to infinity otherwise. Exactly at $p = 1$ the integral turns into $ln b$ , which crawls upward forever — so the harmonic series diverges. That razor-sharp cutoff at $p = 1$ is the punchline.

Connections

Integral test — proof, p-series (index 4.3.7) — the parent note this walkthrough visualises.
Improper integrals — the machinery that computes the ramp's area out to infinity.
Harmonic series — the exact borderline $p = 1$ we watched barely diverge.
Comparison test — once you trust the p-series verdict, use it as a yardstick.
Limit comparison test — same yardstick, softer matching rule.
Riemann zeta function — the true value the sum lands on (which the test never reveals).
n-th term test for divergence — why $a_{n} \to 0$ alone is never enough.

#calculus3 #series/convergence