Worked examples — Integral test — proof, p-series

This deep dive is a case-by-case gym for the integral test. The parent note built the machinery; here we drive it into every corner: each sign of $p$, the exact boundary, degenerate starting points, log-tweaked series, a word problem, and an exam trap. Guess each answer before reading the steps.

Before we start, one reminder of the tool we lean on the most. When we write ${\int }_1^{\infty }f(x)dx={\lim }_{b\to \infty }{\int }_1^{b}f(x)dx,$ that limit-of-an-area is an improper integral — see Improper integrals. It answers the question "does the shaded area under the curve stop growing?" We picked the integral and not just staring at the terms because area is something we can compute exactly with antiderivatives, while an infinite sum of terms is not.

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The scenario matrix

Every problem the integral test throws at you falls into one of these cells. The last column names the example that nails it.

Cell	What changes	Question it tests	Example
A. $p>1$	steep decay	converges?	Ex 1
B. $p<1$	gentle decay	diverges?	Ex 2
C. $p=1$ (harmonic)	exact boundary	the razor edge	Ex 3
D. $0<p<1$ fractional / negative $p$	very slow or growing terms	degenerate decay	Ex 4
E. Log-modified	$\frac{1}{n(\ln n{)}^q}$	second-order boundary	Ex 5
F. Shifted start / not decreasing early	begins at $n=N>1$	does the start matter?	Ex 6
G. Bounding the sum's value	use trapping inequality	numbers, not just fate	Ex 7
H. Word problem	real quantity	model then test	Ex 8
I. Exam twist	disguised p-series	spot the $p$	Ex 9

We now walk each cell. The figure below shows the two decay speeds side by side so you feel why the cutoff sits where it does.

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Example 1 — Cell A: steep decay, $p>1$

1. Identify $f$. Set $f(x)=x^{-3}$, so $a_n=f(n)=1/n^3$. Why this step? The integral test needs a continuous stand-in $f$ for the discrete terms.
2. Check the 3 hypotheses. $f>0$ on $[1,\infty )$; it's continuous (no division by zero for $x\ge 1$); and $f^{\prime }(x)=-3x^{-4}<0$, so decreasing. Why this step? Skipping "decreasing" is the classic mistake — the rectangle comparison only runs one clean direction when $f$ falls.
3. Compute the improper integral. ${\int }_1^{\infty }{x}^{-3}dx={\lim }_{b\to \infty }\frac{x^{-2}}{-2}{|}_1^b={\lim }_{b\to \infty }(\frac{-1}{2b^2}+\frac{1}{2})=\frac{1}{2}.$ Why this step? Here $p=3>1$ makes the exponent $1-p=-2$ negative, so $b^{1-p}\to 0$ — the area stops growing.
4. Conclude. Finite integral ⇒ converges.

Verify: general rule gives ${\int }_1^{\infty }{x}^{-p}=\frac{1}{p-1}=\frac{1}{3-1}=\frac{1}{2}$. ✓ Matches.

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Example 2 — Cell B: gentle decay, $p<1$

1. Rewrite as a power. $\frac{1}{\sqrt{n}}=n^{-1/2}$, so $p=\frac{1}{2}$. Why this step? Recognising the $p$ instantly classifies the series.
2. Integrate $f(x)=x^{-1/2}$. ${\int }_1^{\infty }{x}^{-1/2}dx={\lim }_{b\to \infty }2\sqrt{x}{|}_1^b={\lim }_{b\to \infty }(2\sqrt{b}-2)=\infty .$ Why this step? $1-p=\frac{1}{2}>0$, so $b^{1-p}\to \infty$ — the area never stops.
3. Conclude. Divergent integral ⇒ diverges, even though $a_n\to 0$.

Verify: $p=0.5\le 1$, so the p-series rule predicts divergence. ✓ And this contradicts the tempting "terms $\to 0$" mistake — see n-th term test for divergence (it can only prove divergence, never convergence).

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Example 3 — Cell C: the exact boundary, harmonic series

1. Note $p=1$. The power rule for the antiderivative ($\frac{x^{1-p}}{1-p}$) breaks here because $1-p=0$ divides by zero. Why this step? $p=1$ is genuinely a separate case; the antiderivative of $1/x$ is $\ln x$, not a power.
2. Integrate with the log. ${\int }_1^{\infty }\frac{dx}{x}={\lim }_{b\to \infty }\ln b=\infty .$ Why the logarithm? Because $\int x^{-1}dx=\ln x$ — the one power whose antiderivative is a log. This is the unique reason $p=1$ diverges instead of converging.
3. Conclude. Diverges — this is the Harmonic series.

Verify: $\ln b\to \infty$, and $p=1\ngtr 1$. ✓ Both agree.

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Example 4 — Cell D: degenerate decay ($0<p<1$ fractional, and $p\le 0$)

1. (a) Classify. $p=0.9<1$ ⇒ by the rule, diverges. Why this step? No matter how close to $1$ from below, $1-p>0$ still sends $b^{1-p}\to \infty$.
2. (a) Integral check. ${\int }_1^{\infty }{x}^{-0.9}dx=\frac{x^{0.1}}{0.1}{|}_1^{\infty }=\infty$. Confirms divergence.
3. (b) Degenerate case $p=-2$. Then $a_n=1/n^{-2}=n^2\to \infty$, so terms don't even approach $0$. Why this step? The integral test requires a decreasing $f$. Here $f(x)=x^2$ increases — the test's hypotheses fail, so we don't even reach for it.
4. (b) Use the cheaper tool. By the n-th term test for divergence: $a_n=n^2\nrightarrow 0$ ⇒ diverges immediately.

Verify: (a) $p=0.9\le 1$ ⇒ diverges. ✓ (b) ${\lim }_{n\to \infty }{n}^2=\infty \ne 0$ ⇒ diverges by term test. ✓

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Example 5 — Cell E: the second-order boundary $\frac{1}{n(\ln n{)}^q}$

1. Set up $f$ and start at $n=2$. $f(x)=\frac{1}{x(\ln x{)}^q}$; we start at $2$ because $\ln 1=0$ would divide by zero. Why this step? Convergence never cares about a finite front chunk, so shifting the start to $2$ is free.
2. Substitute $u=\ln x$, $du=dx/x$. ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^q}={\int }_{\ln 2}^{\infty }\frac{du}{u^q}.$ Why the substitution? It converts the messy integrand into a plain p-integral in $u$ — the whole problem collapses to "is $\int u^{-q}du$ finite?".
3. Apply the p-integral rule in $u$. ${\int }^{\infty }{u}^{-q}du$ converges $⟺q>1$. Why this step? We already proved exactly this in the p-series section — now reused on the variable $u$.
4. Conclude. Converges $⟺q>1$. For $q=1$ it's the $\sum \frac{1}{n\ln n}$ divergence from the parent note.

Verify: At $q=2$: ${\int }_{\ln 2}^{\infty }{u}^{-2}du=\frac{1}{\ln 2}$ (finite) ⇒ converges. At $q=1$: $\int u^{-1}du=\infty$ ⇒ diverges. ✓ Numerically $\frac{1}{\ln 2}\approx 1.4427$.

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Example 6 — Cell F: shifted start / not decreasing at first

1. Handle $n=1$ separately. At $n=1$, $\ln 1=0$, so that term is $0$; harmless. Why this step? One finite term never changes convergence.
2. Check eventual decrease. $f(x)=\frac{\ln x}{x^2}$, and $f^{\prime }(x)=\frac{1-2\ln x}{x^3}<0⟺\ln x>\frac{1}{2}⟺x>e^{1/2}\approx \mathrm{1.65.}$ So $f$ decreases for $x\ge 2$. Why this step? The integral test only needs $f$ eventually decreasing — start the test from $n=2$.
3. Integrate by parts, $u=\ln x,dv=x^{-2}dx$: $\int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}-\frac{1}{x}+C.$ Why this step? Products of a log and a power beg for integration by parts.
4. Take the limit from $2$ to $\infty$. As $x\to \infty$, $\frac{\ln x}{x}\to 0$ and $\frac{1}{x}\to 0$, so ${\int }_2^{\infty }\frac{\ln x}{x^2}dx=[0-0]-(-\frac{\ln 2}{2}-\frac{1}{2})=\frac{\ln 2}{2}+\frac{1}{2},$ a finite number ⇒ converges.

Verify: $\frac{\ln 2}{2}+\frac{1}{2}\approx 0.8466$ (finite). ✓ The cutoff $x=e^{1/2}$ satisfies $\ln (e^{1/2})=\frac{1}{2}$. ✓

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Example 7 — Cell G: bounding the value with the trapping inequality

1. Recall the tail-trap. For a convergent decreasing $f$, the remainder $R_N={\sum }_{n=N+1}^{\infty }{a}_n$ obeys ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f.$ Why this step? Same rectangle logic as the parent proof, applied to the leftover after $N$ terms.
2. Plug $f=x^{-2}$, $N=10$. ${\int }_N^{\infty }{x}^{-2}dx=\frac{1}{N}$, so $\frac{1}{11}\le R_{10}\le \frac{1}{10}.$ Why this step? ${\int }_a^{\infty }{x}^{-2}=1/a$ — a clean closed form.
3. Add the known partial sum $S_{10}={\sum }_{n=1}^{10}\frac{1}{n^2}$. Then $S=S_{10}+R_{10}$ gives $S_{10}+\frac{1}{11}\le S\le S_{10}+\frac{1}{10}.$ Why this step? This is the payoff — the integral test traps but never equals the sum (the second classic mistake).

Verify: $S_{10}\approx 1.549768$, so bounds are $[1.640677,1.649768]$; and indeed ${\pi }^2/6\approx 1.644934$ lies inside. ✓

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Example 8 — Cell H: word problem

1. Model. Total ink $={\sum }_{n=1}^{\infty }\frac{1}{n^{1.5}}$ mL — a p-series with $p=1.5$. Why this step? Translate the physical quantity into a series before testing.
2. Classify. $p=1.5>1$ ⇒ converges: the total ink is a finite number of mL.
3. Bound it (how much?). By the tail argument, $S\le a_1+{\int }_1^{\infty }{x}^{-1.5}dx=1+\frac{1}{1.5-1}=1+2=3$ mL. Why this step? Converts an abstract "finite" into a usable units-carrying upper bound of $3$ mL.

Verify: ${\int }_1^{\infty }{x}^{-1.5}dx=\frac{1}{1.5-1}=2$, so the crude bound $1+2=3$ mL holds. (True value $\zeta (1.5)\approx 2.612$ mL — see Riemann zeta function — which is under $3$.) ✓

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Example 9 — Cell I: exam twist (disguised p-series)

1. Find the dominant behaviour. For large $n$, $\frac{n+3}{n^3+n}\approx \frac{n}{n^3}=\frac{1}{n^2}$. Why this step? Leading powers dominate; this hints $p=2$.
2. Confirm rigorously with the Limit comparison test. Compare with $b_n=1/n^2$: ${\lim }_{n\to \infty }\frac{(n+3)/(n^3+n)}{1/n^2}={\lim }_{n\to \infty }\frac{n^2(n+3)}{n^3+n}={\lim }_{n\to \infty }\frac{n^3+3n^2}{n^3+n}=1.$ Why this step? A finite, nonzero limit means both series share the same fate — cleaner than integrating a rational function directly.
3. Conclude. Since $\sum 1/n^2$ converges ($p=2>1$), so does the given series. Converges.

Verify: the limit-comparison ratio limit equals $1$ (finite, nonzero), and the yardstick $p=2>1$ converges. ✓

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Active recall

Recall Which cell, which tool?

• Terms grow ($p\le 0$) ::: n-th term test — one line, diverges.
• Log in the denominator ::: substitute $u=\ln x$, reduces to a p-integral in $u$.
• Rational function of $n$ ::: limit comparison with $1/n^{(\mathrm{deg}\text{denom}-\mathrm{deg}\text{numer})}$.
• Need an actual number bound ::: trapping / tail inequality, never equality.

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Connections

• Integral test — proof, p-series — the parent: proof and the p-series rule.
• Comparison test — bound a messy series above/below by a p-series.
• Limit comparison test — used in Ex 9 to tame the rational function.
• Improper integrals — the limit-of-area engine inside every example.
• Harmonic series — Ex 3, the exact boundary.
• Riemann zeta function — the true values we bounded in Ex 7 and Ex 8.
• n-th term test for divergence — the fast filter in Ex 4.

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