Compare each term with the one before it, or take its $n$th root: if the terms shrink at least as fast as a geometric series the sum converges — and a limit of exactly $1$ tells you nothing.
How would you decide whether $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}$ or $\displaystyle\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2\,5^n}$ converges? The comparison tests of Section 8.4 need a simpler series to compare with, and it is not at all obvious which one to choose. The integral test of Section 8.3 is hopeless — there is no reasonable function $f(x)$ with $f(n)=n!$ to integrate.
Series built from factorials and $n$th powers are exactly the ones that appear in probability (Poisson and binomial models), in algorithms for computing $e$, $\pi$ and $\sin x$, and — in Chapter 9 — in every power series. The tests of this section handle them by asking one question: does the series eventually behave like a convergent geometric series? A geometric series $\sum ar^n$ has a constant ratio $\frac{a_{n+1}}{a_n}=r$ and a constant $n$th root $\sqrt[n]{|ar^n|}\to|r|$. The ratio test and the root test measure these two quantities for any series. We finish with a complete strategy for choosing among all the tests in this chapter.
The Ratio Test
📐 Theorem 8.22 — The ratio test
Let $\sum a_n$ be a series whose terms are non-zero (at least from some index on), and suppose the limit
$$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$
exists or is $+\infty$. (This limit is also written $\rho$.)
If $0\le L\lt1$, the series is absolutely convergent (hence convergent).
If $L\gt1$ or $L=\infty$, the series diverges.
If $L=1$, the test is inconclusive: the series may converge or diverge, and another test is needed.
Why it works. Suppose $L\lt1$ and choose a number $r$ with $L\lt r\lt1$. Because the ratios approach $L$, there is an index $N$ such that, for all $n\ge N$, $|a_{n+1}|\le r\,|a_n|$. Applying this repeatedly, $$|a_{N+1}|\le r|a_N|,\quad|a_{N+2}|\le r^2|a_N|,\quad\dots,\quad|a_{N+k}|\le|a_N|\,r^k.$$ The geometric series $\sum|a_N|r^k$ converges because $0\lt r\lt1$, so $\sum|a_n|$ converges by comparison. If $L\gt1$, then eventually $|a_{n+1}|\gt|a_n|$: the sizes of the terms grow, so $a_n\not\to0$ and the series diverges by the divergence test. When $L=1$ neither argument works — and the examples below show that both outcomes really occur.
✏️ Example 1 — A factorial in the denominator
Does $\displaystyle\sum_{n=1}^{\infty}\frac{2^n}{n!}$ converge?
💡 Show solution▶
Form the ratio $a_n=\dfrac{2^n}{n!}$, so $$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}=\frac{2^{n+1}}{2^n}\cdot\frac{n!}{(n+1)!}=2\cdot\frac{1}{n+1}.$$ The key simplification is $(n+1)!=(n+1)\cdot n!$.
Conclude The series converges (absolutely). The factorial eventually beats any exponential: from $n=3$ on, $a_{n+1}\le\tfrac12a_n$: every term is at most half of the one before. (In Chapter 9 we will see that the sum is $e^2-1\approx6.389$.)
Form the ratio $$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n.$$
Limit $\left(\dfrac{n}{n+1}\right)^n=\dfrac{1}{\left(1+\frac1n\right)^n}\to\dfrac1e$, using the Calculus I limit $\left(1+\frac1n\right)^n\to e$.
Conclude $L=\dfrac1e\approx0.368\lt1$, so the series converges. Interactive 1 shows these ratios settling at $0.368$.
Simplify the factorials $(2n+2)!=(2n+2)(2n+1)\,(2n)!$ and $\big((n+1)!\big)^2=(n+1)^2(n!)^2$. So $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)\,(2n)!}{(n+1)^2(n!)^2\,5^{n+1}}\cdot\frac{(n!)^2\,5^n}{(2n)!}=\frac{(2n+2)(2n+1)}{5(n+1)^2}=\frac{2(2n+1)}{5(n+1)}.$$
Limit Dividing top and bottom by $n$: $L=\displaystyle\lim_{n\to\infty}\frac{2(2+1/n)}{5(1+1/n)}=\frac45\lt1$.
Conclude The series converges. (Its sum turns out to be $\sqrt5-1\approx1.236$.) Note that with $4^n$ instead of $5^n$ the limit would be exactly $1$ and the ratio test could not decide.
⚠️ When $L=1$ the ratio test says nothing
Compare two $p$-series:
$$\sum\frac1n:\quad\frac{a_{n+1}}{a_n}=\frac{n}{n+1}\to1,\qquad\qquad\sum\frac1{n^2}:\quad\frac{a_{n+1}}{a_n}=\frac{n^2}{(n+1)^2}\to1.$$
Both give $L=1$, yet the first diverges and the second converges. In fact every $p$-series, and every series whose terms behave like a power of $n$ (rational functions, roots, logarithms), gives $L=1$. For those series use comparison, limit comparison or the integral test instead.
Other traps. (1) "$L\lt1$" must be about the limit: for $\sum\frac1n$ every ratio $\frac{n}{n+1}$ is less than $1$, but the limit is $1$. (2) Simplify with $(n+1)!=(n+1)\,n!$ — never cancel $(n+1)!$ against $n!$ to leave "$1$". (3) Put the absolute value in: the ratio test proves absolute convergence, so signs are ignored.
Interactive 1 — Watching the Ratios (and Roots) Settle
For the chosen series, the red dots are the ratios $\left|a_{n+1}/a_n\right|$ and the blue dots are the roots $\sqrt[n]{|a_n|}$, plotted against $n$. Both settle to the same limit $L$ (green). Below the orange line $L=1$ the series converges; above it, it diverges; on it — as for $\sum 1/n$ and $\sum 1/n^2$ — the tests cannot tell.
$n$—
ratio $|a_{n+1}/a_n|$—
root $\sqrt[n]{|a_n|}$—
limit $L$—
conclusion—
The Root Test
When the whole term is an $n$th power, such as $\left(\frac{2n+3}{3n+1}\right)^n$, the ratio $\frac{a_{n+1}}{a_n}$ is messy, but the $n$th root strips off the power at once.
📐 Theorem 8.23 — The root test
Let $\sum a_n$ be a series and suppose the limit
$$L=\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}|a_n|^{1/n}$$
exists or is $+\infty$.
If $0\le L\lt1$, the series is absolutely convergent.
If $L\gt1$ or $L=\infty$, the series diverges.
If $L=1$, the test is inconclusive.
Why it works. If $L\lt r\lt1$, then eventually $|a_n|^{1/n}\le r$, i.e. $|a_n|\le r^n$, and $\sum r^n$ is a convergent geometric series. If $L\gt1$, eventually $|a_n|^{1/n}\gt1$, so $|a_n|\gt1$ and the terms cannot tend to $0$. For the $p$-series, $\sqrt[n]{1/n^p}=\big(n^{1/n}\big)^{-p}\to1$, so the root test is just as helpless there as the ratio test.
Behind the last line is Stirling's formula $n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, which also explains why $\sqrt[n]{n!/n^n}\to\frac1e$ (Example 2).
Conclude The series converges by the root test. (Numerically its sum is about $0.1046$.)
Compare Do not confuse this with $\displaystyle\sum\left(1-\frac1n\right)^{n}$: there the terms tend to $e^{-1}\ne0$, so that series diverges by the divergence test — and its root test gives $L=\lim\left(1-\frac1n\right)=1$, which is inconclusive. The exponent $n^2$ makes all the difference.
A Strategy for Testing Series
We now have nine tools. The hardest part of a series problem is often deciding which one to use. No recipe works every time, but the following order, based on the form of $a_n$, settles the vast majority of first-year problems quickly.
🧭 Strategy for testing $\sum a_n$
Divergence test. If you can see that $a_n\not\to0$, the series diverges. Stop. (If $a_n\to0$, you have learned nothing yet — go on.)
Special series. Geometric $\sum ar^{n}$: converges iff $|r|\lt1$. $p$-series $\sum\frac1{n^p}$: converges iff $p\gt1$. Telescoping: write out $S_N$.
Factorials, or products of exponentials and powers ($n!$, $c^n$, $n^k c^n$): try the ratio test.
The whole term is an $n$th power $(b_n)^n$: try the root test.
Alternating $\sum(-1)^nb_n$: try the alternating series test; then test $\sum b_n$ to decide between absolute and conditional convergence.
Algebraic / rational expressions (powers and roots of $n$), or terms like a known series: comparison or limit comparison with a $p$-series or geometric series, keeping the dominant terms.
$a_n=f(n)$ with $f$ positive, decreasing and easy to integrate (often involving $\ln n$): the integral test.
Test
Use when $a_n$ looks like…
Conclusion
Divergence
terms with a non-zero limit
only divergence
Geometric / $p$-series
$ar^n$, $\ 1/n^p$
$|r|\lt1$; $\ p\gt1$
Ratio
$n!$, $\ c^n$, $\ n^kc^n$
$L\lt1$ abs. conv.; $\ L\gt1$ div.
Root
$(b_n)^n$
$L\lt1$ abs. conv.; $\ L\gt1$ div.
Alternating series
$(-1)^nb_n$, $\ b_n\downarrow0$
convergence + error $\le b_{N+1}$
(Limit) comparison
rational / algebraic in $n$
same behaviour as the comparison series
Integral
$f(n)$, $f$ positive decreasing
same behaviour as $\int_1^\infty f$
Interactive 2 — The Test-Chooser Flowchart
Pick a series and press Next step (or ▶ Play). The flowchart asks the strategy questions in order, highlights the path taken, and stops at the first test that settles the series. The working appears below the chart.
series—
step—
test used—
verdict—
✏️ Example 6 — Choosing the test
Decide which test to use, and determine convergence or divergence: (a) $\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{n^4+2}$ (b) $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n\,n^2}{3^n}$ (c) $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$ (d) $\displaystyle\sum_{n=1}^{\infty}n\sin\frac1n$
💡 Show solution▶
(a) Rational → limit comparison $a_n\to0$, so step 1 says nothing. The dominant terms give $\dfrac{n^3}{n^4}=\dfrac1n$. With $b_n=\dfrac1n$: $\dfrac{a_n}{b_n}=\dfrac{n^4}{n^4+2}\to1$, a positive finite number, and $\sum\frac1n$ diverges. So the series diverges. (The ratio test would give $L=1$ — useless here.)
(b) Power × exponential → ratio $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^2}{3^{n+1}}\cdot\dfrac{3^n}{n^2}=\dfrac13\left(1+\dfrac1n\right)^2\to\dfrac13\lt1$. Absolutely convergent. (The AST also works, but the ratio test gives the stronger conclusion in one step.)
(c) Logarithm over a power → comparison Since $\ln x\lt\sqrt x$ for all $x\gt0$ (the function $\sqrt x-\ln x$ has its minimum $2-\ln4\gt0$ at $x=4$), $$0\le\frac{\ln n}{n^2}\lt\frac{\sqrt n}{n^2}=\frac1{n^{3/2}},$$ and $\sum n^{-3/2}$ is a convergent $p$-series. Converges by comparison. (The integral test also works.)
(d) Check the limit first With $x=\frac1n\to0^+$: $n\sin\dfrac1n=\dfrac{\sin x}{x}\to1\ne0$. Diverges by the divergence test — no other test is needed.
Exercises
I
Use the ratio test to determine whether each series converges absolutely or diverges. If the test is inconclusive, say so and settle the question with another test.
10 $\dfrac{n+1}{(n+1)^2+1}\cdot\dfrac{n^2+1}{n}\to1$: inconclusive. Limit comparison with $\frac1n$: $\dfrac{n/(n^2+1)}{1/n}=\dfrac{n^2}{n^2+1}\to1$, and $\sum\frac1n$ diverges, so the series diverges.
Mixed practice — choose a suitable test, name it, and decide whether each series converges or diverges. For series with negative terms, say whether convergence is absolute or conditional.
2Comparison: $0\lt\dfrac1{n^2+5n}\lt\dfrac1{n^2}$ and $\sum\frac1{n^2}$ converges. Converges.
3AST: $b_n=\dfrac1{\sqrt{n+3}}$ is positive, decreasing, $\to0$, so it converges. But $\sum\frac1{\sqrt{n+3}}$ diverges (limit comparison with $\frac1{\sqrt n}$, ratio $\to1$). Conditionally convergent.
4Comparison with a geometric series: $\dfrac{3^n}{4^n+1}\lt\left(\dfrac34\right)^n$ and $\sum\left(\frac34\right)^n$ converges. Converges.
5Ratio test: $\dfrac{(n+1)e^{-(n+1)^2}}{n\,e^{-n^2}}=\dfrac{n+1}{n}\,e^{-(2n+1)}\to0$. Converges. (Or the integral test: $\int_1^\infty xe^{-x^2}dx=\frac1{2e}$.)
6Integral test: $f(x)=\dfrac1{x(\ln x)^2}$ is positive and decreasing for $x\ge2$; with $u=\ln x$, $\displaystyle\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\lim_{t\to\infty}\left[-\frac1{\ln x}\right]_2^t=\frac1{\ln2}$, finite. Converges.
12 Ratio and root tests both give $L=1$. Limit comparison with $\frac1n$: $\dfrac{1/n^{1+1/n}}{1/n}=n^{-1/n}=\dfrac{1}{\sqrt[n]{n}}\to1$, and $\sum\frac1n$ diverges. Diverges — even though every exponent $1+\frac1n$ is bigger than $1$!
IV
Concepts and applications.
1Applied (telecommunications). A call centre receives on average $\lambda$ calls per minute, and the probability of exactly $n$ calls in a minute is $p_n=e^{-\lambda}\dfrac{\lambda^n}{n!}$ ($n=0,1,2,\dots$). Use the ratio test to show that $\sum p_n$ and the mean $\sum n\,p_n$ converge for every $\lambda\gt0$. Given that $\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}=e^\lambda$, find the mean number of calls when $\lambda=4$.
2 For which $x\gt0$ does $\displaystyle\sum_{n=1}^{\infty}n\,x^n$ converge?
3 Explain why the ratio test can never decide whether a $p$-series $\sum\frac1{n^p}$ converges.
4 Let $a_n=2^{-n+(-1)^n}$. Show that $\frac{a_{n+1}}{a_n}$ has no limit (so the ratio test cannot be applied) but that the root test proves $\sum a_n$ converges.
5Applied (queueing). At a busy airport check-in desk, the probability that $n$ passengers are waiting is modelled by $p_n=(1-\rho)\rho^n$, $n=0,1,2,\dots$, where $\rho=0.8$ is the desk's utilisation. Show by the ratio test that the mean queue length $\sum n\,p_n$ is finite, and compute it using $\displaystyle\sum_{n=1}^{\infty}n\,x^n=\frac{x}{(1-x)^2}$ for $0\lt x\lt1$.
💡 Show solutions▶
1 $\dfrac{p_{n+1}}{p_n}=\dfrac{\lambda}{n+1}\to0$ and $\dfrac{(n+1)p_{n+1}}{n\,p_n}=\dfrac{\lambda}{n}\to0$, so both series converge for every $\lambda$. Mean: $\displaystyle\sum_{n=1}^{\infty}n\,e^{-\lambda}\frac{\lambda^n}{n!}=\lambda e^{-\lambda}\sum_{n=1}^{\infty}\frac{\lambda^{n-1}}{(n-1)!}=\lambda e^{-\lambda}e^{\lambda}=\lambda$. For $\lambda=4$: 4 calls per minute.
2 $\dfrac{(n+1)x^{n+1}}{n\,x^n}=\dfrac{n+1}{n}\,x\to x$. Converges for $0\lt x\lt1$, diverges for $x\gt1$. At $x=1$ the test is inconclusive, but the terms are $n\not\to0$, so it diverges. Converges exactly for $0\lt x\lt1$.
3 $\dfrac{a_{n+1}}{a_n}=\left(\dfrac{n}{n+1}\right)^p\to1^p=1$ for every $p$, so the test always lands in the inconclusive case — even though $p$-series converge for $p\gt1$ and diverge for $p\le1$.
4 $\dfrac{a_{n+1}}{a_n}=2^{-1+(-1)^{n+1}-(-1)^n}=2^{-1-2(-1)^n}$, which is $2^{-3}=\tfrac18$ for even $n$ and $2^{1}=2$ for odd $n$: no limit. But $\sqrt[n]{a_n}=2^{-1+(-1)^n/n}\to2^{-1}=\tfrac12\lt1$, so the series converges (its sum is $1$).
5 $\dfrac{(n+1)p_{n+1}}{n\,p_n}=\dfrac{n+1}{n}\,\rho\to\rho=0.8\lt1$: converges. Mean $=(1-\rho)\dfrac{\rho}{(1-\rho)^2}=\dfrac{\rho}{1-\rho}=\dfrac{0.8}{0.2}=$ 4 passengers.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How many correct digits of $\pi$ does each new term of a "fast" series add — and can the ratio test predict it?
Why it matters. In 1914 the Indian mathematician Srinivasa Ramanujan published series for $1/\pi$ that converge astonishingly fast. Their descendants (the Chudnovsky brothers' series, 1988) are used for record computations of $\pi$ to trillions of digits, which in turn are used to stress-test supercomputer hardware. The ratio test is exactly the tool that measures their speed.
What to do.
Ramanujan's series is $\displaystyle\frac1\pi=\frac{2\sqrt2}{9801}\sum_{k=0}^{\infty}t_k$ with $t_k=\dfrac{(4k)!\,(1103+26390k)}{(k!)^4\,396^{4k}}$. Simplify $\dfrac{t_{k+1}}{t_k}$ and show that it tends to $L=\dfrac{4^4}{396^4}$.
Explain why, if $|t_{k+1}/t_k|\approx L$, each extra term multiplies the error by about $L$, so it adds about $-\log_{10}L$ correct digits. Compute this number.
With a calculator or spreadsheet, compute $\pi$ from one term, then from two terms. Count the correct digits and compare with your prediction.
Compare with the alternating series $4\left(1-\frac13+\frac15-\cdots\right)$ of Section 8.5. How many of its terms would give the same accuracy as one Ramanujan term?
Tools: Calculator, spreadsheet or Python (use high-precision arithmetic for step 3).
Hand in: A half-page note with the limit $L$, the predicted digits per term, the computed approximations and a comparison of the two series.
Going further. The Chudnovsky series has ratio limit $L=\dfrac{1728}{640320^3}$. How many digits per term does it give? Search for how record $\pi$ computations are organised today.
Instructor note: $L=256/396^4\approx1.04\times10^{-8}$, i.e. about $7.98$ digits per term. One term gives $\pi\approx3.14159273$ (error $7.6\times10^{-8}$); two terms give an error of about $6\times10^{-16}$. The Leibniz series needs about $1.3\times10^{7}$ terms for an error of $7.6\times10^{-8}$. Chudnovsky: $L\approx6.58\times10^{-15}$, about $14.18$ digits per term.
🧪 Self-check quiz
0/5 correct
Q1. For $\sum\frac{5^n}{n!}$ the ratio test gives $L=$
Q2. If the ratio test gives $L=1$, then the series
Q3. The most efficient test for $\sum\left(\frac{3n+1}{4n-1}\right)^n$ is
Q4. For $\sum\frac{n!}{n^n}$, $\lim\frac{a_{n+1}}{a_n}=$
Q5. Which statement about $\sum\frac{n}{n+1}$ is correct?