Integral test — proof, p-series

Tag: #calculus3 #series/convergence

The big question (WHY)

We have a series $\sum a_n$ and we want to know: does the infinite sum settle on a finite number, or blow up? Adding infinitely many terms by hand is impossible. The integral test is a trick: replace the discrete sum (a staircase of rectangles) with a continuous integral (an area under a curve) that we can compute.

The test itself (WHAT)

The three conditions matter: positive (so areas don't cancel), decreasing (so the rectangle comparison goes one clean direction), continuous (so the integral exists).

Deriving it from scratch (HOW)

We build the inequality that traps the sum. Picture $y=f(x)$ falling to the right.

Step 1 — Upper-sum rectangles (left endpoints). On each interval $[n, n+1]$, since $f$ decreases, the largest value is at the left, $f(n)$, and the smallest is at the right, $f(n+1)$. Draw a rectangle of width $1$ and height $f(n)$ (left endpoint): it sits above the curve, so its area is the upper sum. Why this step? The left-endpoint height $f(n)$ overestimates the area on $[n,n+1]$, giving us an upper bound on the integral.

$\int_{n}^{n+1} f(x)\,dx \;\le\; f(n).$

Sum from $n=1$ to $N$: $\int_{1}^{N+1} f(x)\,dx \;\le\; \sum_{n=1}^{N} f(n) = S_N.$

Step 2 — Lower-sum rectangles (right endpoints). Now use height $f(n+1)$ (right endpoint), which sits below the curve, so its area is the lower sum: $f(n+1) \;\le\; \int_{n}^{n+1} f(x)\,dx.$ Sum from $n=1$ to $N-1$: $S_N - f(1) = \sum_{n=2}^{N} f(n) \;\le\; \int_{1}^{N} f(x)\,dx.$

Step 3 — Squeeze. Combine: $\int_{1}^{N+1} f(x)\,dx \;\le\; S_N \;\le\; f(1) + \int_{1}^{N} f(x)\,dx.$

Step 4 — Read off the conclusion.

• If $\int_1^\infty f$ converges (finite), the right side stays bounded, so $S_N$ is increasing and bounded above ⇒ $\sum a_n$ converges.
• If $\int_1^\infty f$ diverges ($\to\infty$), the left side $\to\infty$, dragging $S_N\to\infty$ ⇒ $\sum a_n$ diverges. $\blacksquare$

The payoff: the p-series (80/20)

Apply the test with $f(x)=x^{-p}$ (positive, continuous, decreasing for $p>0$). Compute the integral.

Case $p\ne 1$: $\int_{1}^{\infty} x^{-p}\,dx = \lim_{b\to\infty}\frac{x^{1-p}}{1-p}\Big|_1^b = \lim_{b\to\infty}\frac{b^{1-p}-1}{1-p}.$

• If $p>1$: $1-p<0$, so $b^{1-p}\to 0$, integral $= \dfrac{1}{p-1}$ (finite) ⇒ converges.
• If $p<1$: $1-p>0$, so $b^{1-p}\to\infty$ ⇒ diverges.

Case $p=1$ (harmonic series): $\int_1^\infty \frac{dx}{x} = \lim_{b\to\infty}\ln b = \infty \;\Rightarrow\; \textbf{diverges}.$

Worked examples

Steel-manning the classic mistakes

Active recall

Connections

• Comparison test — uses p-series as the standard "yardstick".
• Limit comparison test — compare against $1/n^p$.
• Improper integrals — the engine inside the proof.
• Harmonic series — the boundary case $p=1$.
• Riemann zeta function — $\zeta(p)=\sum 1/n^p$ for $p>1$.
• n-th term test for divergence — the necessary (not sufficient) condition.

Concept Map

Hinglish (regional understanding)