Exercises — Integral test — proof, p-series

The rule: $\sum 1/n^p$ converges $⟺p>1$.

• (a) $\frac{1}{n^3}=\frac{1}{n^3}$, so $p=3>1$ ⇒ converges.
• (b) $\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$, so $p=\frac{1}{2}<1$ ⇒ diverges.
• (c) $\frac{1}{n}=\frac{1}{n^1}$, so $p=1$ ⇒ diverges (this is the Harmonic series, the exact boundary case).

Check the three hypotheses restated at the top of the page.

• (a) Positive ✓, continuous ✓, and $f^{\prime }(x)=-2x^{-3}<0$ so decreasing ✓ ⇒ legal.
• (b) $\sin x$ swings between $-1$ and $1$: not positive, not decreasing ⇒ illegal (areas would cancel).
• (c) $\frac{1}{x}-2$ is negative for $x>\frac{1}{2}$ ⇒ not positive on $[1,\infty )$ ⇒ illegal.

WHAT: set $f(x)=\frac{1}{x^2+1}$. WHY legal (all three hypotheses): positive ✓, continuous ✓; $f^{\prime }(x)=\frac{-2x}{(x^2+1{)}^2}<0$ for $x\ge 1$ so decreasing ✓. ${\int }_1^{\infty }\frac{dx}{x^2+1}={\lim }_{b\to \infty }[\arctan x{]}_1^b={\lim }_{b\to \infty }(\arctan b-\arctan 1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}.$ WHY $\arctan$? The antiderivative of $\frac{1}{x^2+1}$ is $\arctan x$ — this is the one integral whose value is an angle, and $\arctan b\to \frac{\pi }{2}$ as $b\to \infty$. The integral is finite ($\pi /4\approx 0.785$) ⇒ the series converges.

WHAT: $f(x)=x{e}^{-x^2}$, positive on $[1,\infty )$; $f^{\prime }(x)=e^{-x^2}(1-2x^2)<0$ for $x\ge 1$ ⇒ decreasing ✓. WHY the substitution $u=x^2$: the factor $x$ out front is exactly $\frac{1}{2}du$ where $du=2xdx$ — it lets the whole integrand collapse to $e^{-u}$. ${\int }_1^{\infty }x{e}^{-x^2}dx\stackrel{u=x^2}{=}\frac{1}{2}{\int }_1^{\infty }{e}^{-u}du=\frac{1}{2}[-e^{-u}{]}_1^{\infty }=\frac{1}{2}{e}^{-1}=\frac{1}{2e}.$ Finite ($\approx 0.184$) ⇒ the series converges.

WHERE it decreases: $f^{\prime }(x)=\frac{1-\ln x}{x^2}$. This is $<0$ when $\ln x>1$, i.e. $x>e\approx 2.718$. So $f$ increases up to $x=e$, then decreases.

What to see in figure s02: the blue curve $y=\ln x/x$ climbs from $x=1$, reaches a single peak (the red dot) right at the yellow dashed line $x=e$, then falls forever after. Left of the peak it rises (green label), so the integral-test rectangle comparison would go the wrong way there; right of the peak it falls (red label), where the test is legal.

WHY that's fine: the integral test only needs $f$ eventually decreasing — start from $N=3$ (the tail decides convergence; finitely many front terms can't change the fate). ${\int }_3^{\infty }\frac{\ln x}{x}dx\stackrel{u=\ln x}{=}{\int }_{\ln 3}^{\infty }udu={\lim }_{b\to \infty }\frac{u^2}{2}{|}_{\ln 3}^b=\infty .$ WHY $u=\ln x$: $du=dx/x$ absorbs the $\frac{1}{x}$, leaving the pure $u$, which grows without bound. Integral diverges ⇒ the series diverges.

WHAT: $f(x)=\frac{1}{x(\ln x{)}^p}$ on $[2,\infty )$. WHY positive: for $x\ge 2$, $\ln x>0$ so both factors in the denominator are positive ✓. WHY continuous: it's a quotient of continuous functions with denominator never zero on $[2,\infty )$ ✓. WHY decreasing: write $g(x)=x(\ln x{)}^p$ (the denominator); then $g^{\prime }(x)=(\ln x{)}^p+x\cdot p(\ln x{)}^{p-1}\cdot \frac{1}{x}=(\ln x{)}^{p-1}(\ln x+p)>0\text{for }x\ge 2,$ since $\ln x>0$ and $\ln x+p>0$. A growing positive denominator means $f=1/g$ is decreasing ✓. WHY $u=\ln x$: $du=dx/x$ turns the integrand into a pure p-integral in $u$. ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^p}\stackrel{u=\ln x}{=}{\int }_{\ln 2}^{\infty }\frac{du}{u^p}.$ Now it is a p-integral in $u$, whose convergence we already know:

• $p>1$: $\int u^{-p}du$ converges ⇒ series converges.
• $p\le 1$: diverges ⇒ series diverges. So the answer is the same razor cutoff as the p-series: convergence $⟺p>1$. (At $p=1$ we recover the parent's Example 3, which diverges.)

WHY not integrate: $\frac{1}{\sqrt{x^3+1}}$ has no clean antiderivative. Instead compare with a p-series whose fate we know from the integral test. For all $n\ge 1$: $\sqrt{n^3+1}>\sqrt{n^3}=n^{3/2}$, so $\frac{1}{\sqrt{n^3+1}}<\frac{1}{n^{3/2}}.$ WHY this bound helps: $\sum \frac{1}{n^{3/2}}$ is a p-series with $p=\frac{3}{2}>1$, which the integral test proves converges. A smaller positive series under a convergent one also converges (Comparison test). ⇒ converges. (Cross-check with Limit comparison test: $\lim \frac{1/\sqrt{n^3+1}}{1/n^{3/2}}=1$, a finite nonzero limit, so same fate.)

Recall the trapping inequality with $S_N={\sum }_{n=1}^{N}f(n)$ the $N$-term partial sum: ${\int }_1^{N+1}f\le S_N\le f(1)+{\int }_1^{N}f.$ Here $f(x)=x^{-2}$, $f(1)=1$, and ${\int }_1^{\infty }{x}^{-2}dx=1$ (parent Example 1). Let $N\to \infty$ so $S_N\to S$: $\underset{=1}{\underbrace{{\int }_1^{\infty }\frac{dx}{x^2}}}\le S\le \underset{=1}{\underbrace{f(1)}}+\underset{=1}{\underbrace{{\int }_1^{\infty }\frac{dx}{x^2}}}.$ So $1\le S\le 2$. WHY this is the point of the test: it bounds the sum without evaluating it. The true value is ${\pi }^2/6\approx 1.645$, which indeed lands in $[1,2]$ — never equal to either integral (that's the classic mistake). Tightening: bound the tail ${\sum }_{n=2}^{\infty }$ by ${\int }_1^{\infty }{x}^{-2}=1$, giving $S\le 1+1=2$ and $S\ge 1+{\int }_2^{\infty }{x}^{-2}=1+\frac{1}{2}=1.5$.

The idea (same rectangle geometry as the top of the page, shifted to the tail). Over $[n,n+1]$, decreasing $f$ gives $f(n+1)\le {\int }_n^{n+1}f\le f(n).$

What to see in figure s03: the blue curve is $y=1/x^2$ on the tail $x\ge 3$. The red dashed rectangles use left-endpoint height $f(n)$ and poke above the curve (their sum overshoots ⇒ upper bound); the green rectangles use right-endpoint height $f(n+1)$ and sit under it (their sum undershoots ⇒ lower bound). The remainder $R_N$ is trapped between the two shaded areas.

Upper bound for $R_N$: using the left inequality $f(n+1)\le {\int }_n^{n+1}f$, sum $n=N,N+1,\dots$: $R_N={\sum }_{n=N+1}^{\infty }f(n)={\sum }_{k=N}^{\infty }f(k+1)\le {\sum }_{k=N}^{\infty }{\int }_k^{k+1}f={\int }_N^{\infty }f.$ Lower bound for $R_N$: using the right inequality ${\int }_n^{n+1}f\le f(n)$, sum $n=N+1,N+2,\dots$: ${\int }_{N+1}^{\infty }f={\sum }_{n=N+1}^{\infty }{\int }_n^{n+1}f\le {\sum }_{n=N+1}^{\infty }f(n)=R_N.$ Combine: ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f$. $■$

Apply to $\sum 1/n^2$, $N=3$: ${\int }_N^{\infty }{x}^{-2}dx=\frac{1}{N}$, so $\frac{1}{4}\le R_3\le \frac{1}{3},\text{i.e. }0.25\le R_3\le 0.333\dots$ (True remainder $R_3={\pi }^2/6-1-\frac{1}{4}-\frac{1}{9}\approx 0.2838$ — inside the band.)

Partial sum $S_3=1+\frac{1}{4}+\frac{1}{9}=\frac{49}{36}\approx 1.3611$. From Problem 5.1, $\frac{1}{4}\le R_3\le \frac{1}{3}$. So $S_3+\frac{1}{4}\le S\le S_3+\frac{1}{3},1.6111\le S\le \mathrm{1.6944.}$ Midpoint estimate: $S\approx \frac{1.6111+1.6944}{2}=1.6528$, with maximum error $\frac{1.6944-1.6111}{2}\approx 0.0417$. The true value ${\pi }^2/6\approx 1.6449$ is inside $[1.6111,1.6944]$ and within $0.008$ of our midpoint — far better than the crude $[1,2]$ of Problem 4.2. WHY it's better: we computed a few exact terms and only estimated the tail with integrals, where the rectangle error is small.

• Define the $N$-term partial sum $S_N$. ::: $S_N={\sum }_{n=1}^{N}f(n)$, the running total of the first $N$ terms.
• State the trapping inequality. ::: ${\int }_1^{N+1}f\le S_N\le f(1)+{\int }_1^{N}f$.
• $p$ for $\frac{1}{2n}$ and its fate? ::: $p=1$, diverges (constant $\frac{1}{2}$ irrelevant).
• Antiderivative used for $\sum \frac{1}{n^2+1}$? ::: $\arctan x$; integral $=\pi /4$, converges.
• Where does $\frac{\ln x}{x}$ start decreasing? ::: For $x>e$; test from $N=3$.
• Cutoff for $\sum \frac{1}{n(\ln n{)}^p}$? ::: Converges $⟺p>1$.
• Remainder bound for a convergent integral-test series? ::: ${\int }_{N+1}^{\infty }f\le R_N\le {\int }_N^{\infty }f$.