35. The ratio and root tests
Comparison tests need a known series to compare against, and the integral test needs an antiderivative. Neither works well on series built from factorials or terms raised to the nth power — n! has no useful integral form, and comparing \frac{2^n}{n!} against a p-series is a dead end. This closing lesson of the module supplies two tests purpose- built for exactly those shapes, and they happen to be the two tests Module 8 leans on most.
The ratio test
L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
- If L<1: \sum a_n converges absolutely.
- If L>1 (including L=\infty): \sum a_n diverges.
- If L=1: inconclusive — the test says nothing.
Why this works. If \left|\frac{a_{n+1}}{a_n}\right|\to L<1, then for large n each term is roughly L times the one before — behaving like a geometric series with ratio L<1 (§7.3), which converges. Formally, pick any r with L<r<1; eventually |a_{n+1}|\le r|a_n|, so the tail is bounded by a convergent geometric series, and direct comparison (§7.7) finishes the argument. The L>1 case is the mirror image: terms eventually grow, so they can't shrink to zero, and the divergence test (§7.5) applies.
Why L=1 tells you nothing: both \sum\frac1n (diverges) and \sum\frac1{n^2} (converges) give L=1 under the ratio test — it simply can't distinguish between series whose consecutive-term ratio approaches 1 without settling the question of how the terms behave overall. When L=1, fall back to comparison or the integral test instead.
Where this test dominates: any series with n! or a constant raised to the nth power, because both simplify beautifully under the ratio. \frac{(n+1)!}{n!}=n+1 exactly (nearly everything cancels), and \frac{r^{n+1}}{r^n}=r exactly — this is precisely the algebra the comparison tests couldn't offer a shortcut for.
The root test
L=\lim_{n\to\infty}\sqrt[n]{|a_n|}
Same three-way conclusion as the ratio test: L<1 converges absolutely, L>1 diverges, L=1 inconclusive.
Why this works: if \sqrt[n]{|a_n|}\to L, then eventually |a_n|\approx L^n — again behaving like a geometric series with ratio L, by the same comparison argument as the ratio test.
Where this test dominates: series where the entire term is raised to the nth power, like \left(\frac{n}{2n+1}\right)^n — the nth root simply undoes the outer exponent, leaving something far simpler behind. The ratio test on a term like this would leave an awkward \left(\frac{n+1}{n}\right)^{n+1}-style expression to simplify; the root test dissolves it in one step.
Choosing between them: ratio test for factorials and products of fixed-base powers; root test for anything already wrapped in an outer nth power. Both tests, when they apply at all, always agree on the verdict — they're proving the same underlying "eventually geometric" fact through different algebra.
Doing it in Python
The ratio test on \sum\frac{2^n}{n!} — factorial in the denominator guarantees convergence no matter how fast the numerator's exponential grows, echoing §1.6's "factorials beat exponentials" hierarchy:
import sympy as sp
n = sp.Symbol('n', positive=True, integer=True)
a_n = 2**n / sp.factorial(n)
ratio = sp.limit(a_n.subs(n, n + 1) / a_n, n, sp.oo)
print(f"lim |a_(n+1)/a_n| = {ratio}")
print(f"L < 1, so the series converges absolutely")
The ratio test on \sum\frac{n!}{n^n} — landing on \frac1e, a direct sighting of §7.0's \left(1+\frac1n\right)^n\to e limit, hiding inside the simplified ratio:
import sympy as sp
n = sp.Symbol('n', positive=True, integer=True)
a_n = sp.factorial(n) / n**n
ratio_expr = sp.simplify(a_n.subs(n, n + 1) / a_n)
print(f"a_(n+1)/a_n simplifies to: {ratio_expr}")
ratio = sp.limit(ratio_expr, n, sp.oo)
print(f"limit = {ratio} (this is 1/e, since (n/(n+1))^n -> 1/e)")
print(f"L = 1/e < 1, converges")
The root test on \sum\left(\frac n{2n+1}\right)^n — the outer power dissolving in one step:
import sympy as sp
n = sp.Symbol('n', positive=True, integer=True)
a_n = (n / (2*n + 1))**n
root = sp.limit(sp.Abs(a_n)**(1/n), n, sp.oo)
print(f"lim |a_n|^(1/n) = {root}")
print(f"L < 1, converges absolutely")
Worked example
Determine whether \displaystyle\sum_{n=1}^\infty\dfrac{n^2}{3^n} converges, using the ratio test.
\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^2/3^{n+1}}{n^2/3^n}=\frac{(n+1)^2}{n^2}\cdot\frac{3^n}{3^{n+1}}=\frac{(n+1)^2}{n^2}\cdot\frac13
L=\lim_{n\to\infty}\frac{(n+1)^2}{n^2}\cdot\frac13=1\cdot\frac13=\frac13
since \frac{(n+1)^2}{n^2}=\left(1+\frac1n\right)^2\to1^2=1.
\boxed{\frac13<1\ \Longrightarrow\ \text{converges absolutely}}
Sanity check. The polynomial factor n^2 grows without bound, but the exponential 3^n in the denominator eventually dominates any polynomial — exactly §1.6's "exponentials beat polynomials" hierarchy, here confirmed rigorously rather than just cited. The ratio test's constant limit of \frac13 makes sense: for large n, the \left(1+\frac1n\right)^2 factor is negligibly close to 1, so consecutive terms really are shrinking by close to a factor of 3 each step — an almost-geometric series with ratio close to \frac13, which is exactly the intuition behind why the test works at all. ✓
Your turn
1. Use the ratio test on \displaystyle\sum_{n=1}^\infty\frac{n!}{2^n}.
2. Use the root test on \displaystyle\sum_{n=1}^\infty\left(\frac{3n+1}{4n}\right)^n.
3. True or false: the ratio test proves \displaystyle\sum_{n=1}^\infty\frac1n diverges, since \left|\frac{a_{n+1}}{a_n}\right|=\frac n{n+1}\to1.
Solutions
1.
\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!/2^{n+1}}{n!/2^n}=(n+1)\cdot\frac12
L=\lim_{n\to\infty}\frac{n+1}2=\infty
\boxed{L>1\ \Longrightarrow\ \text{diverges}}
Factorial growth eventually overwhelms any fixed exponential base — the reverse of the worked example, and again matching §1.6's growth hierarchy.
2.
L=\lim_{n\to\infty}\sqrt[n]{\left(\frac{3n+1}{4n}\right)^n}=\lim_{n\to\infty}\frac{3n+1}{4n}=\frac34
\boxed{\frac34<1\ \Longrightarrow\ \text{converges}}
3. False. L=1 is the inconclusive case — the ratio test proves nothing here, neither convergence nor divergence. (The harmonic series does diverge, as §7.5 proved with Oresme's grouping argument, but the ratio test itself is simply the wrong tool to establish it — a good reminder that L=1 always means "try a different test," never "the series diverges.")
Check yourself in code
Using the ratio test, compute L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| for a_n=\frac{2^n}{n!}, and using the root test, compute L=\lim_{n\to\infty}\sqrt[n]{|a_n|} for a_n=\left(\frac n{2n+1}\right)^n.
Print exactly this:
ratio test, 2^n/n!: L = 0
root test, (n/(2n+1))^n: L = 1/2
import sympy as sp
n = sp.Symbol('n', positive=True, integer=True)
a1 = 2**n / sp.factorial(n)
L1 = sp.limit(a1.subs(n, n + 1) / a1, n, sp.oo)
print("ratio test, 2^n/n!: L = ...")
a2 = (n / (2*n + 1))**n
L2 = sp.limit(sp.Abs(a2)**(1/n), n, sp.oo)
print("root test, (n/(2n+1))^n: L = ...")
import sympy as sp
n = sp.Symbol('n', positive=True, integer=True)
a1 = 2**n / sp.factorial(n)
L1 = sp.limit(a1.subs(n, n + 1) / a1, n, sp.oo)
print(f"ratio test, 2^n/n!: L = {L1}")
a2 = (n / (2*n + 1))**n
L2 = sp.limit(sp.Abs(a2)**(1/n), n, sp.oo)
print(f"root test, (n/(2n+1))^n: L = {L2}")
Both tests reduce a series to an "eventually geometric" comparison: the ratio test via \left|\frac{a_{n+1}}{a_n}\right|\to L, the root test via \sqrt[n]{|a_n|}\to L, converging absolutely when L<1 and diverging when L>1, with L=1 settling nothing either way. The ratio test is the natural choice whenever factorials or fixed-base powers appear, since the algebra collapses cleanly; the root test wins when the whole term already sits inside an nth power. Both will reappear immediately in Module 8, where the ratio test becomes the standard tool for finding a power series' radius of convergence.
That closes this module's tour of every classical convergence test — divergence, integral, comparison, alternating, ratio, and root — the complete toolkit against which nearly any series you'll meet can be checked. Next: power series, where the coefficients themselves depend on n in a structured way, and a convergent series becomes a genuine representation of a function.