Chapter 8 · Sequences and Series

8.4 Comparison Tests

Decide whether a positive series converges by measuring it against a series you already know — a geometric series or a $p$-series — either term by term or through the limit of the ratio $a_n/b_n$.

The integral test of Section 8.3 is powerful, but it needs an integral we can actually evaluate and a function that is eventually decreasing. Try it on $\displaystyle\sum\frac{1}{3^n+n}$ or $\displaystyle\sum\frac{2+\sin n}{n^2}$ and you get stuck immediately: the first integral has no elementary antiderivative, and the second function is not decreasing at all.

Yet for large $n$ the first series obviously “looks like” the geometric series $\sum(1/3)^n$, and the second is squeezed under $\sum 3/n^2$. This section turns that instinct into proof. We compare an unknown series with a benchmark whose behaviour we already know — mostly the geometric series $\sum r^n$ (converges exactly when $|r|\lt 1$) and the $p$-series $\sum 1/n^p$ (converges exactly when $p\gt 1$). Throughout, the series have positive (or non-negative) terms.

The Direct Comparison Test

When every $a_n\ge0$, the partial sums $S_N=a_1+\cdots+a_N$ can only go up. By the Monotone Convergence Theorem (Section 8.1) such a series converges exactly when its partial sums are bounded above. So if a series with larger terms has a finite sum, that sum is a ceiling for ours; and if a series with smaller terms already runs off to infinity, ours is dragged along with it.

📐 Theorem 8.15 — The direct comparison test
Suppose there is an integer $N_0$ such that $$0\le a_n\le b_n\qquad\text{for all }n\ge N_0.$$
  1. If $\sum b_n$ converges, then $\sum a_n$ converges.
  2. If $\sum a_n$ diverges, then $\sum b_n$ diverges.
Reduce The first $N_0-1$ terms do not affect convergence, so we may assume $0\le a_n\le b_n$ for every $n\ge1$. Let $S_N=\sum_{n=1}^N a_n$ and $T_N=\sum_{n=1}^N b_n$. Both sequences are increasing, and $S_N\le T_N$.
Part 1 If $\sum b_n=T$ converges, then $S_N\le T_N\le T$ for every $N$. So $(S_N)$ is increasing and bounded above, hence convergent.
Part 2 If $\sum a_n$ diverges, its increasing partial sums are unbounded: $S_N\to\infty$. Since $T_N\ge S_N$, also $T_N\to\infty$. ∎
⚠️ Analysis — the two directions that tell you nothing

1. Smaller than a divergent series: no conclusion. $\frac1{n^2}\le\frac1n$ and $\sum\frac1n$ diverges, but $\sum\frac1{n^2}$ converges; while $\frac1{2n}\le\frac1n$ and $\sum\frac1{2n}$ diverges.

2. Larger than a convergent series: no conclusion. $\frac1n\ge\frac1{n^2}$ (diverges) and $\frac2{n^2}\ge\frac1{n^2}$ (converges).

3. The inequality only has to hold eventually — from some $N_0$ on — and the terms must be non-negative. For instance $\frac{-1}{n}\le\frac{1}{n^2}$, but $\sum\frac{-1}{n}$ diverges.

Memory aid: to prove convergence, find a bigger convergent series; to prove divergence, find a smaller divergent series.

✏️ Example 1 — Proving convergence by comparison
Show that both series converge. (a) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+3n+1}$   (b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^n+n}$
(a) Compare For $n\ge1$ the denominator satisfies $n^2+3n+1\gt n^2$, so $$0\lt\frac{1}{n^2+3n+1}\lt\frac{1}{n^2}.$$
(a) Conclude $\sum\frac1{n^2}$ is a $p$-series with $p=2\gt 1$, so it converges. The given series is term-by-term smaller, so it converges (its sum is about $0.5463$).
(b) Compare $3^n+n\gt 3^n$, so $0\lt\dfrac{1}{3^n+n}\lt\dfrac{1}{3^n}=\left(\dfrac13\right)^n$.
(b) Conclude $\sum\left(\frac13\right)^n$ is geometric with $r=\tfrac13$, $|r|\lt 1$, so it converges (to $\tfrac12$). Hence the given series converges, and its sum is less than $\tfrac12$ (it is about $0.3921$).
✏️ Example 2 — Proving divergence by comparison
Show that both series diverge. (a) $\displaystyle\sum_{n=3}^{\infty}\frac{\ln n}{n}$   (b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}}$
(a) For $n\ge3\gt e$ we have $\ln n\gt 1$, so $\dfrac{\ln n}{n}\gt\dfrac1n\gt 0$. The harmonic series $\sum\frac1n$ diverges, and our terms are larger, so $\displaystyle\sum_{n=3}^\infty\frac{\ln n}{n}$ diverges. (The inequality fails at $n=1,2$ — that does not matter.)
(b) A first try $\dfrac{1}{\sqrt{n+1}}\lt\dfrac1{\sqrt n}$ — smaller than a divergent series, which tells us nothing. We need an inequality the other way.
(b) Fix it For $n\ge1$, $n+1\le2n$, so $\sqrt{n+1}\le\sqrt2\,\sqrt n$ and $$\frac{1}{\sqrt{n+1}}\ge\frac{1}{\sqrt2}\cdot\frac{1}{n^{1/2}}.$$ The right side is a constant multiple of a divergent $p$-series ($p=\tfrac12$), so the given series diverges.
Lesson Finding the inequality in the useful direction can take some fiddling. The limit comparison test below removes that fiddling.

The Limit Comparison Test

Consider $\displaystyle\sum\frac{1}{2^n-1}$. It obviously behaves like $\sum\frac1{2^n}$, but $\frac{1}{2^n-1}\gt\frac1{2^n}$ — bigger than a convergent series, the useless direction. What really matters is not which terms are bigger, but that for large $n$ the two are almost proportional: their ratio tends to a positive finite number.

📐 Theorem 8.16 — The limit comparison test
Suppose $a_n\gt 0$ and $b_n\gt 0$ for all $n\ge N_0$, and let $$L=\lim_{n\to\infty}\frac{a_n}{b_n}.$$
  1. If $0\lt L\lt\infty$, then $\sum a_n$ and $\sum b_n$ both converge or both diverge.
  2. If $L=0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
  3. If $L=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
Case 1 Since $a_n/b_n\to L\gt 0$, eventually $\tfrac{L}{2}\lt\dfrac{a_n}{b_n}\lt\tfrac{3L}{2}$, i.e. $\tfrac L2\,b_n\lt a_n\lt\tfrac{3L}{2}\,b_n$. If $\sum b_n$ converges, so does $\sum\frac{3L}{2}b_n$, and the right inequality with direct comparison gives convergence of $\sum a_n$. If $\sum b_n$ diverges, so does $\sum\frac L2 b_n$, and the left inequality gives divergence of $\sum a_n$.
Case 2 $L=0$: eventually $\dfrac{a_n}{b_n}\lt 1$, so $0\lt a_n\lt b_n$; direct comparison with the convergent $\sum b_n$.
Case 3 $L=\infty$: eventually $\dfrac{a_n}{b_n}\gt 1$, so $a_n\gt b_n\gt 0$; direct comparison with the divergent $\sum b_n$. ∎
⚠️ Analysis — when the limit comparison test is silent

1. $L=0$ with $\sum b_n$ divergent says nothing: $a_n$ is much smaller than a divergent series. ($a_n=\frac1{n^2}$, $b_n=\frac1n$: $L=0$ and $\sum a_n$ converges; $a_n=\frac1n$, $b_n=\frac{1}{\sqrt n}$: $L=0$ and $\sum a_n$ diverges.)

2. $L=\infty$ with $\sum b_n$ convergent says nothing either — $a_n$ is much bigger than a convergent series.

3. If $\lim a_n/b_n$ does not exist the test does not apply (see Example 6(b)). Try a direct comparison instead.

In practice, if you land in case 1 or 2 of this box, your comparison series was badly chosen: pick a $b_n$ that matches the size of $a_n$ more closely.

Interactive 1 — Partial Sums Side by Side, and the Ratio That Decides

Left: the partial sums of the series under test $\sum a_n$ (red) and of the benchmark $\sum b_n$ (blue) for $n$ from $1$ to $1000$ (horizontal axis $\log_{10}n$). Right: the ratio $a_n/b_n$ settling towards $L$ (dashed). When $0\lt L\lt\infty$ the two sums share the same fate; watch the special cases $L=0$ and $L=\infty$ too.

partial sums $S_n$ against $\log_{10}n$
ratio $a_n/b_n$ against $\log_{10}n$
$n$—
$\sum_{k\le n}a_k$—
$\sum_{k\le n}b_k$—
ratio $a_n/b_n$—
✏️ Example 3 — A rational function of $n$
Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{3n+1}{n^3-2n+4}$ converges.
Positivity The numerator is positive, and $n^3-2n+4$ equals $3,8,25,\dots$ for $n=1,2,3$ and increases for $n\ge1$, so $a_n\gt 0$.
Choose $b_n$ For large $n$, $a_n\approx\dfrac{3n}{n^3}=\dfrac{3}{n^2}$. Take $b_n=\dfrac1{n^2}$. (Direct comparison is awkward here: the $-2n$ in the denominator makes $a_n$ bigger than $\frac{3}{n^2}$ for many $n$.)
Limit $$L=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(3n+1)n^2}{n^3-2n+4}=\lim_{n\to\infty}\frac{3+1/n}{1-2/n^2+4/n^3}=3.$$
Conclude $0\lt 3\lt\infty$ and $\sum\frac1{n^2}$ converges ($p=2$), so $\displaystyle\sum\frac{3n+1}{n^3-2n+4}$ converges.
✏️ Example 4 — A trigonometric series
Determine whether $\displaystyle\sum_{n=1}^{\infty}\sin\frac1n$ converges.
Positivity For $n\ge1$, $0\lt\frac1n\le1\lt\pi$, so $\sin\frac1n\gt 0$. The divergence test is inconclusive since $\sin\frac1n\to0$.
Choose $b_n$ For small angles $\sin\theta\approx\theta$, so $\sin\frac1n\approx\frac1n$. Take $b_n=\dfrac1n$.
Limit With $\theta=\frac1n\to0^+$: $\displaystyle L=\lim_{n\to\infty}\frac{\sin(1/n)}{1/n}=\lim_{\theta\to0^+}\frac{\sin\theta}{\theta}=1.$
Conclude $0\lt 1\lt\infty$ and the harmonic series diverges, so $\displaystyle\sum\sin\frac1n$ diverges. (By contrast, $\sum\sin\frac{1}{n^2}$ converges by the same argument with $b_n=\frac1{n^2}$.)
✏️ Example 5 — Logarithms and the case $L=0$
Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ converges, using the limit comparison test.
Two bad choices With $b_n=\frac1{n^2}$ (convergent): $\frac{a_n}{b_n}=\ln n\to\infty$ — $L=\infty$ with a convergent $\sum b_n$, no conclusion. With $b_n=\frac1n$ (divergent): $\frac{a_n}{b_n}=\frac{\ln n}{n}\to0$ — $L=0$ with a divergent $\sum b_n$, no conclusion. The truth lies between $\frac1{n^2}$ and $\frac1n$.
Good choice Logarithms grow more slowly than any positive power, so give away a little of the power: take $b_n=\dfrac{1}{n^{3/2}}$, a convergent $p$-series.
Limit $\displaystyle L=\lim_{n\to\infty}\frac{\ln n/n^2}{1/n^{3/2}}=\lim_{n\to\infty}\frac{\ln n}{\sqrt n}=\lim_{x\to\infty}\frac{1/x}{1/(2\sqrt x)}=\lim_{x\to\infty}\frac{2}{\sqrt x}=0$ (L'Hôpital).
Conclude $L=0$ and $\sum b_n$ converges, so $\displaystyle\sum\frac{\ln n}{n^2}$ converges — agreeing with the integral test in Example 4 of Section 8.3, with much less work.

Choosing a Comparison Series

🧭 Finding $b_n$ from the dominant terms
  1. Check $\lim a_n$ first. If it is not $0$, the series diverges and you are done.
  2. Keep only the dominant term of the numerator and of the denominator — the term that grows fastest as $n\to\infty$. Use the growth ladder $$\ln n\ \ll\ n^{p}\ (p\gt 0)\ \ll\ r^n\ (r\gt 1)\ \ll\ n!$$ and, for powers, the highest power wins. Roots act on powers: $\sqrt{n^3+n}\approx n^{3/2}$.
  3. Simplify the ratio of dominant terms and drop constant factors: this is $b_n$ — usually $\frac1{n^p}$ or $r^n$.
  4. Compute $L=\lim a_n/b_n$ (divide top and bottom by the dominant power). With the dominant-term choice you should get $0\lt L\lt\infty$.
  5. Logarithms: if $\ln n$ survives, compare with a slightly different power ($n^{-3/2}$ instead of $n^{-2}$, say) and use the case $L=0$ or $L=\infty$.
  6. Oscillating factors such as $\sin n$, $\cos n$, $(-1)^n$ inside a bounded expression: bound them ($1\le2+\sin n\le3$) and use direct comparison.
Interactive 2 — Find the Dominant Term, Step by Step

Choose a series and press Next step. The graph is on log–log axes ($\log_{10}a_n$ against $\log_{10}n$), where a power $n^{-p}$ is a straight line of slope $-p$ (the geometric example uses $n$ itself across the bottom instead). The red curve is $a_n$; once the comparison series appears (blue), notice that the two run parallel for large $n$: their vertical gap tends to $\log_{10}L$.

    largest $n$ shown—
    $a_n$—
    $b_n$—
    $a_n/b_n$—
    ✏️ Example 6 — When one test fails and the other works
    (a) Show that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n-1}$ converges: explain why direct comparison with $\sum\frac1{2^n}$ fails, and use the limit comparison test instead.
    (b) Show that $\displaystyle\sum_{n=1}^{\infty}\frac{2+\sin n}{n^2}$ converges: explain why the limit comparison test with $\sum\frac1{n^2}$ fails, and use direct comparison instead.
    (a) DCT fails $2^n-1\lt 2^n$, so $\dfrac1{2^n-1}\gt\dfrac1{2^n}$: our series is bigger than a convergent one — no conclusion.
    (a) LCT works Both sequences are positive and $$L=\lim_{n\to\infty}\frac{1/(2^n-1)}{1/2^n}=\lim_{n\to\infty}\frac{2^n}{2^n-1}=\lim_{n\to\infty}\frac{1}{1-2^{-n}}=1.$$ Since $\sum\left(\frac12\right)^n$ is a convergent geometric series, $\displaystyle\sum\frac1{2^n-1}$ converges (sum $\approx1.6067$). Remark: a direct comparison can be rescued — $2^n-1\ge2^{n-1}$ gives $\frac1{2^n-1}\le\frac{2}{2^n}$ — but you have to think of it.
    (b) LCT fails $\dfrac{a_n}{b_n}=\dfrac{(2+\sin n)/n^2}{1/n^2}=2+\sin n$, which keeps oscillating between about $1$ and $3$ and has no limit. The limit comparison test cannot be applied.
    (b) DCT works Since $-1\le\sin n\le1$, we have $0\lt\dfrac{2+\sin n}{n^2}\le\dfrac{3}{n^2}$, and $\sum\frac3{n^2}$ converges. So the series converges (sum $\approx4.30$).

    Exercises

    I
    Use the direct comparison test. In each case state the comparison series, the inequality (and from which $n$ it holds), and your conclusion.
    • 1 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3+4}$
    • 2 $\displaystyle\sum_{n=2}^{\infty}\frac{1}{\sqrt n-1}$
    • 3 $\displaystyle\sum_{n=1}^{\infty}\frac{\sin^2n}{n^2}$
    • 4 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^n+2}$
    • 5 $\displaystyle\sum_{n=1}^{\infty}\frac{n+1}{n^2}$
    • 6 $\displaystyle\sum_{n=1}^{\infty}\frac{2^n}{3^n+1}$
    • 7 $\displaystyle\sum_{n=2}^{\infty}\frac{1}{\ln n}$
    • 8 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n!}$ (hint: $n!\ge2^{n-1}$)
    • 9 $\displaystyle\sum_{n=1}^{\infty}\frac{\tan^{-1}n}{n^{3/2}}$
    1 $0\lt\dfrac1{n^3+4}\lt\dfrac1{n^3}$ for $n\ge1$; $\sum\frac1{n^3}$ converges ($p=3$). Converges.
    2 For $n\ge2$, $0\lt\sqrt n-1\lt\sqrt n$, so $\dfrac1{\sqrt n-1}\gt\dfrac1{\sqrt n}$; $\sum\frac1{\sqrt n}$ diverges ($p=\tfrac12$). Diverges.
    3 $0\le\sin^2n\le1$, so $0\le\dfrac{\sin^2n}{n^2}\le\dfrac1{n^2}$; $\sum\frac1{n^2}$ converges. Converges.
    4 $0\lt\dfrac1{3^n+2}\lt\left(\dfrac13\right)^n$; geometric with $r=\tfrac13$. Converges.
    5 $\dfrac{n+1}{n^2}=\dfrac1n+\dfrac1{n^2}\gt\dfrac1n\gt 0$; harmonic series diverges. Diverges.
    6 $0\lt\dfrac{2^n}{3^n+1}\lt\dfrac{2^n}{3^n}=\left(\dfrac23\right)^n$; geometric with $r=\tfrac23$. Converges.
    7 $0\lt\ln n\lt n$ for $n\ge2$, so $\dfrac1{\ln n}\gt\dfrac1n$; harmonic diverges. Diverges.
    8 $n!=1\cdot2\cdot3\cdots n\ge1\cdot2\cdot2\cdots2=2^{n-1}$, so $0\lt\dfrac1{n!}\le\dfrac{1}{2^{n-1}}=2\left(\dfrac12\right)^n$; a convergent geometric series. Converges (to $e-1$, as Chapter 9 will show).
    9 $0\lt\tan^{-1}n\lt\dfrac\pi2$ for $n\ge1$, so $0\lt\dfrac{\tan^{-1}n}{n^{3/2}}\lt\dfrac{\pi}{2}\cdot\dfrac1{n^{3/2}}$; a constant times a convergent $p$-series ($p=\tfrac32$). Converges.
    II
    Use the limit comparison test. State $b_n$, compute $L$, and conclude.
    • 1 $\displaystyle\sum_{n=1}^{\infty}\frac{n+2}{n^3+1}$
    • 2 $\displaystyle\sum_{n=1}^{\infty}\frac{2n-1}{n^2+3n}$
    • 3 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+5n}}$
    • 4 $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt n+1}{n^2+2}$
    • 5 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n-n}$
    • 6 $\displaystyle\sum_{n=2}^{\infty}\frac{n^2-n}{n^4+2}$
    • 7 $\displaystyle\sum_{n=1}^{\infty}\tan\frac1n$
    • 8 $\displaystyle\sum_{n=1}^{\infty}\left(1-\cos\frac1n\right)$
    • 9 $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\frac1{n^2}\right)$
    Check: watch $a_n/b_n$ approach your value of $L$
    1 $b_n=\dfrac1{n^2}$: $L=\displaystyle\lim\frac{n^3+2n^2}{n^3+1}=1$. $\sum\frac1{n^2}$ converges, so the series converges.
    2 $b_n=\dfrac1n$: $L=\displaystyle\lim\frac{2n^2-n}{n^2+3n}=2$. Harmonic diverges, so the series diverges.
    3 $b_n=\dfrac1n$: $L=\displaystyle\lim\frac{n}{\sqrt{n^2+5n}}=\lim\frac{1}{\sqrt{1+5/n}}=1$. Diverges.
    4 Dominant terms $\frac{\sqrt n}{n^2}=n^{-3/2}$, so $b_n=\dfrac1{n^{3/2}}$: $L=\displaystyle\lim\frac{n^2+n^{3/2}}{n^2+2}=1$. $p=\tfrac32\gt 1$: converges.
    5 $2^n-n\ge1\gt 0$ for $n\ge1$. $b_n=\dfrac1{2^n}$: $L=\displaystyle\lim\frac{2^n}{2^n-n}=\lim\frac{1}{1-n/2^n}=1$ (since $n/2^n\to0$). Geometric $r=\tfrac12$: converges.
    6 Terms positive for $n\ge2$. $b_n=\dfrac1{n^2}$: $L=\displaystyle\lim\frac{n^4-n^3}{n^4+2}=1$. Converges.
    7 $\tan\frac1n\gt 0$ since $0\lt\frac1n\le1\lt\frac\pi2$. $b_n=\dfrac1n$: $L=\displaystyle\lim_{\theta\to0^+}\frac{\tan\theta}{\theta}=1$. Diverges.
    8 $1-\cos\theta\approx\frac{\theta^2}{2}$ for small $\theta$, so take $b_n=\dfrac1{n^2}$: with $\theta=\frac1n$, $L=\displaystyle\lim_{\theta\to0^+}\frac{1-\cos\theta}{\theta^2}=\lim_{\theta\to0^+}\frac{\sin\theta}{2\theta}=\frac12$ (L'Hôpital). Converges.
    9 $\ln(1+x)\approx x$ for small $x$, so $b_n=\dfrac1{n^2}$: with $x=\frac1{n^2}$, $L=\displaystyle\lim_{x\to0^+}\frac{\ln(1+x)}{x}=1$. Converges.
    III
    Mixed practice. Decide whether each series converges or diverges, using any test from Sections 8.3–8.4. Justify your choice of test.
    • 1 $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$
    • 2 $\displaystyle\sum_{n=2}^{\infty}\frac{1}{\sqrt n\,\ln n}$
    • 3 $\displaystyle\sum_{n=1}^{\infty}\frac{(\ln n)^2}{n^{3/2}}$
    • 4 $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{n^2+1}$
    • 5 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{1+1/n}}$ (careful — this is not a $p$-series)
    • 6 $\displaystyle\sum_{n=1}^{\infty}\frac{2+(-1)^n}{n^{3/2}}$
    • 7 $\displaystyle\sum_{n=1}^{\infty}e^{-\sqrt n}$
    • 8 Challenge. $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}$ (hint: show $\frac{n!}{n^n}\le\frac{2}{n^2}$ for $n\ge2$)
    1 LCT with $b_n=\frac1{n^2}$: $\frac{a_n}{b_n}=\frac{\ln n}{n}\to0$ and $\sum\frac1{n^2}$ converges (case $L=0$). Converges. (Or directly: $\ln n\lt n$, so $a_n\lt\frac1{n^2}$.)
    2 LCT with $b_n=\frac1n$: $\frac{a_n}{b_n}=\frac{\sqrt n}{\ln n}\to\infty$ (L'Hôpital: $\frac{1/(2\sqrt x)}{1/x}=\frac{\sqrt x}{2}\to\infty$) and $\sum\frac1n$ diverges (case $L=\infty$). Diverges.
    3 LCT with $b_n=\frac{1}{n^{5/4}}$ (convergent, $p=\tfrac54$): $\frac{a_n}{b_n}=\frac{(\ln n)^2}{n^{1/4}}\to0$ (two applications of L'Hôpital). Case $L=0$: converges.
    4 $\frac{n^2}{n^2+1}\to1\ne0$. Diverges by the divergence test — no comparison needed.
    5 $a_n=\frac{1}{n\cdot n^{1/n}}$. LCT with $b_n=\frac1n$: $\frac{a_n}{b_n}=n^{-1/n}=e^{-(\ln n)/n}\to e^0=1$. Since $\sum\frac1n$ diverges, the series diverges — the exponent $1+\frac1n$ is bigger than $1$ for every $n$, but it is not bounded away from $1$.
    6 $1\le2+(-1)^n\le3$, so $0\lt a_n\le\frac{3}{n^{3/2}}$, a convergent $p$-series multiple. Converges (direct comparison; LCT with $\frac1{n^{3/2}}$ fails because $2+(-1)^n$ has no limit).
    7 LCT with $b_n=\frac1{n^2}$: $\frac{a_n}{b_n}=\frac{n^2}{e^{\sqrt n}}$; with $u=\sqrt n$ this is $\frac{u^4}{e^u}\to0$ (exponentials beat powers). Case $L=0$: converges.
    8 For $n\ge2$: $\dfrac{n!}{n^n}=\dfrac1n\cdot\dfrac2n\cdot\dfrac3n\cdots\dfrac nn\le\dfrac1n\cdot\dfrac2n\cdot1\cdots1=\dfrac{2}{n^2}$, because every factor $\frac kn$ with $3\le k\le n$ is at most $1$. Since $\sum\frac{2}{n^2}$ converges, the series converges by direct comparison.
    IV
    Concepts and applications.
    • 1 True or false? If $0\le a_n\le b_n$ and $\sum a_n$ converges, then $\sum b_n$ converges.
    • 2 True or false? If $a_n\gt 0$ and $\sum a_n$ converges, then $\sum a_n^2$ converges.
    • 3 True or false? If $a_n,b_n\gt 0$, $\lim\frac{a_n}{b_n}=0$ and $\sum a_n$ converges, then $\sum b_n$ converges.
    • 4 For which $p\gt 0$ does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^p+n}$ converge?
    • 5 Applied — an infinite resistor chain. An engineer models a long transmission line as infinitely many resistors in series; the $n$-th has resistance $R_n=\dfrac{n+5}{n^3+2}$ ohms. (a) Show that the total resistance $\sum R_n$ is finite. (b) Using $n+5\le6n$ and $n^3+2\gt n^3$, together with $\sum\frac1{n^2}\le2$ (from Section 8.3), find an upper bound for the total resistance.
    • 6 Applied — pollution from a closed landfill. In year $n$ after closure a landfill leaks $P_n=\dfrac{40}{n\sqrt{n+3}}$ tonnes of a pollutant. (a) Show that the total amount ever leaked is finite and give an upper bound. (b) Would the total still be finite if instead $P_n=\dfrac{40}{\sqrt{n+3}}$ tonnes?
    1 False Being bigger than a convergent series proves nothing: $a_n=\frac1{n^2}\le\frac1n=b_n$, and $\sum\frac1n$ diverges.
    2 True Since $\sum a_n$ converges, $a_n\to0$, so eventually $0\lt a_n\lt 1$ and then $a_n^2\lt a_n$. Direct comparison gives convergence of $\sum a_n^2$.
    3 False $a_n=\frac1{n^2}$, $b_n=\frac1n$: $\frac{a_n}{b_n}=\frac1n\to0$ and $\sum a_n$ converges, but $\sum b_n$ diverges.
    4 If $p\gt 1$ the dominant term is $n^p$: LCT with $\frac1{n^p}$ gives $L=\lim\frac{n^p}{n^p+n}=1$, so it converges. If $p=1$ the terms are $\frac1{2n}$: diverges. If $0\lt p\lt 1$ the dominant term is $n$: LCT with $\frac1n$ gives $L=\lim\frac{n}{n^p+n}=1$, so it diverges. Converges exactly when $p\gt 1$.
    5 (a) LCT with $b_n=\frac1{n^2}$: $L=\lim\frac{n^3+5n^2}{n^3+2}=1$, so $\sum R_n$ converges — the total resistance is finite. (b) For $n\ge1$, $R_n\le\dfrac{6n}{n^3}=\dfrac{6}{n^2}$, so $\sum R_n\le6\sum\frac1{n^2}\le6\cdot2=$ $12$ Ω. (Using $\sum\frac1{n^2}=\frac{\pi^2}{6}$ sharpens this to $\pi^2\approx9.87$ Ω; numerically the total is about $3.45$ Ω.)
    6 (a) $\sqrt{n+3}\gt\sqrt n$, so $0\lt P_n\lt\dfrac{40}{n^{3/2}}$, and $\sum\frac{1}{n^{3/2}}$ converges ($p=\tfrac32$). By the integral-test bound of Section 8.3, $\sum\frac1{n^{3/2}}\le1+\int_1^\infty x^{-3/2}dx=1+2=3$, so the total is at most $40\cdot3=$ $120$ tonnes (numerically about $72.6$ tonnes). (b) No: LCT with $b_n=\frac{1}{\sqrt n}$ gives $L=\lim\frac{40\sqrt n}{\sqrt{n+3}}=40$, and $\sum\frac1{\sqrt n}$ diverges, so the total would be infinite — the leak would never effectively stop.
    🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

    Research question. A person takes in one unit of a substance every day (a medicine, a trace metal in food, a micro-pollutant). A fraction $R(k)$ of each day's intake is still in the body $k$ days later. Does the body burden stay bounded over a lifetime, or keep growing?

    Why it matters. Pharmacologists, toxicologists and radiation-protection agencies (for example when modelling bone-seeking elements) must know whether repeated small doses reach a safe plateau. Long-term burden after $N$ days is $\sum_{k=1}^{N}R(k)$ — a partial sum — and the plateau exists exactly when the series converges.

    What to do.

    1. Consider three retention models: (A) $R(k)=e^{-0.1k}$ (exponential clearance); (B) $R(k)=(1+0.05k)^{-2}$; (C) $R(k)=(1+0.05k)^{-0.8}$ (both “power-law” clearance).
    2. Use the comparison tests of this section to decide which of $\sum R(k)$ converge. (Hint: geometric for A; limit comparison with $1/k^2$ and $1/k^{0.8}$ for B and C.)
    3. In a spreadsheet compute the burden after $1$, $10$ and $100$ years ($N=365$, $3650$, $36\,500$).
    4. For model B, use the remainder estimate of Section 8.3 to estimate the long-run plateau.

    Tools: Spreadsheet or Python.

    Hand in: A half-page note with your table, the convergence verdict for each model with the test used, and one sentence on why the tail of $R(k)$, not its early values, decides the long-term risk.

    Going further. Look up a published retention model for a substance relevant to your country (fluoride in drinking water, lead, or a long-lived medicine) and decide which of the three behaviours it resembles.

    Instructor note: A converges to $e^{-0.1}/(1-e^{-0.1})\approx9.51$ units within weeks; B has $L=400$ against $1/k^2$ and plateaus near $19.51$ (about $18.47$ after one year); C has $L=20^{0.8}\approx11.0$ against the divergent $\sum 1/k^{0.8}$ and grows without bound — about $80$, $183$ and $349$ units after $1$, $10$ and $100$ years.

    🧪 Self-check quiz

    0/5 correct

    Q1. Since $0\lt\dfrac{1}{n^2+7}\lt\dfrac1{n^2}$, the series $\displaystyle\sum\frac{1}{n^2+7}$

    Q2. The best comparison series for $\displaystyle\sum\frac{4n^3+n}{2n^5-1}$ is $\sum b_n$ with $b_n=$

    Q3. You find $\lim a_n/b_n=0$ and $\sum b_n$ diverges. Then $\sum a_n$

    Q4. Why does direct comparison with $\sum\frac1{2^n}$ fail to settle $\displaystyle\sum\frac{1}{2^n-1}$?

    Q5. The series $\displaystyle\sum_{n=1}^{\infty}\sin\frac{1}{n^2}$

    ← 8.3 The Divergence and Integral Tests8.5 Alternating Series →