The Ratio and Root Tests
Both tests ask one question: does the series shrink like a geometric series? The ratio test measures \(L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\); the root test measures \(L = \lim_{n\to\infty}\sqrt[n]{|a_n|}\). Either way: \(L < 1\) means the series converges absolutely, \(L > 1\) means it diverges, and \(L = 1\) tells you nothing.
By the end you'll be able to compute the ratio or root limit \(L\) for a series, read off the verdict from the three cases, choose the friendlier of the two tests for a given term, and explain why absolute convergence implies convergence.
Predict: for \(\sum n/2^n\), does the ratio \(\left|a_{n+1}/a_n\right|\) settle above or below 1? Slide n to check — then switch series to see all three verdicts.
The chart tracks two sequences of measurements as \(n\) grows: the ratio \(\left|a_{n+1}/a_n\right|\) and the root \(\sqrt[n]{|a_n|}\). Early values wobble, but both home in on the same limit \(L\). The verdict only depends on where \(L\) lands: the green zone (\(L < 1\)) converges absolutely, the red zone (\(L > 1\)) diverges, and the amber line at exactly 1 is inconclusive. Hover or focus the dots for exact values.
At n = 10: ratio |aₙ₊₁/aₙ| = 0.550 · root ⁿ√|aₙ| = 0.629 → limit L = 0.500: converges absolutely (L < 1)
The ratio and root tests decide absolute convergence by measuring geometric-like growth: a limit below 1 means the tail shrinks at least geometrically fast, a limit above 1 means the terms eventually grow, and a limit of exactly 1 decides nothing.
Let \(L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\) (ratio test) or \(L = \lim_{n\to\infty}\sqrt[n]{|a_n|}\) (root test). If \(L < 1\), \(\sum a_n\) converges absolutely — meaning \(\sum |a_n|\) converges — and absolute convergence implies convergence, so \(\sum a_n\) converges too. If \(L > 1\) (including \(L = \infty\)), the terms eventually grow, so \(a_n \not\to 0\) and the series diverges. If \(L = 1\), the test is inconclusive: both the divergent harmonic series and the convergent \(\sum 1/n^2\) give \(L = 1\). Because both tests examine \(|a_n|\), they work on series with mixed or complicated signs, not just positive-term ones.
In practice: write \(\left|\frac{a_{n+1}}{a_n}\right|\), cancel everything that telescopes, and take the limit. Reach for the ratio test when the term has factorials or products that collapse under \(a_{n+1}/a_n\) (like \(3^n/n!\)); reach for the root test when the term is already an \(n\)th power (like \(\left(\frac{2n+1}{3n}\right)^n\)), because the \(n\)th root cancels the exponent in one step. Both give the same three-way verdict — pick whichever simplifies more. And when \(L = 1\), don't force it: switch to a comparison, integral, or alternating-series argument. The comparison tests you met last lesson match a series against a geometric or p-series benchmark you supply; the ratio test builds that geometric benchmark directly out of the series' own consecutive terms.
\(\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}\): \(L = \lim_{n\to\infty}\left|\frac{(n+1)/2^{n+1}}{n/2^n}\right| = \lim_{n\to\infty}\frac{n+1}{2n} = \frac12 < 1\), so the series converges absolutely (and hence converges). This is exactly the default series in the chart above — watch the blue ratio curve settle onto \(L = 0.5\), deep inside the green zone.
\(\displaystyle\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^{n}\): the term is an \(n\)th power, so the root test is the natural pick. \(L = \lim_{n\to\infty}\sqrt[n]{\left(\frac{n}{2n+1}\right)^{n}} = \lim_{n\to\infty}\frac{n}{2n+1} = \frac12 < 1\), so the series converges absolutely. One root, one limit, done — no messy consecutive-term quotient needed.
Try \(\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n!}\) — the factorial says ratio test. Set up the quotient: \(\left|\frac{a_{n+1}}{a_n}\right| = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1}\). Now finish it: \(L = \lim_{n\to\infty}\frac{3}{n+1} = \) ____, so the series ____.
Reveal the answer
\(L = 0 < 1\), so the series converges absolutely. The factorial in the denominator crushes the exponential: every extra term multiplies by \(3/(n+1)\), which shrinks toward 0 — faster-than-geometric decay. Any \(L\) strictly below 1 gives the same verdict; \(L = 0\) is just the easiest case.
More info — why \(L < 1\) forces convergence (the geometric idea)
If the consecutive ratios settle near \(L < 1\), then beyond some index every term is at most \(r\) times the previous one, for some fixed \(r\) between \(L\) and 1. Chaining that inequality, the tail is term-by-term smaller than a geometric series with ratio \(r < 1\) — which converges. That's a comparison-test argument in disguise: the ratio test automates the comparison against a geometric benchmark, just like you compared against geometric and p-series by hand last lesson. Conversely, if \(L > 1\) the terms eventually grow, so they can't approach 0 and the series diverges by the divergence test. At \(L = 1\) the geometric picture blurs — terms shrink, but maybe too slowly (\(\sum 1/n\)) or fast enough (\(\sum 1/n^2\)) — which is why the test goes silent there. Paul's Online Math Notes (linked in Dive deeper below) walks through both proofs.
Check your understanding
Apply the ratio test to \(\sum_{n=1}^{\infty} \dfrac{n!}{100^n}\). What does it conclude?
Use the root test on \(\sum_{n=1}^{\infty} \left(\dfrac{2n+1}{3n}\right)^{n}\). What does it conclude?
The ratio test gives \(L = 1\) for both \(\sum \frac{1}{n}\) and \(\sum \frac{1}{n^2}\). What does that tell you about these two series?
The ratio test on \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) gives \(L = 1/2\). In comparison-test language, what is the ratio test really saying about this series' tail?
Recap
- Ratio test: \(L = \lim\left|\frac{a_{n+1}}{a_n}\right|\). Root test: \(L = \lim\sqrt[n]{|a_n|}\). Same three cases: \(L < 1\) converges absolutely, \(L > 1\) diverges, \(L = 1\) inconclusive.
- Absolute convergence (\(\sum |a_n|\) converges) implies convergence of \(\sum a_n\) — and both tests always test \(|a_n|\), so they handle mixed signs.
- \(L = 1\) genuinely decides nothing: \(\sum 1/n\) diverges and \(\sum 1/n^2\) converges, yet both give \(L = 1\). Switch tests.
- Choose ratio for factorials and telescoping products; choose root when the term is already an \(n\)th power. Both measure the same geometric-like decay.
Dive deeper
- Paul's Online Math Notes — Ratio Test Confirm the three-case ratio test and the inconclusive L=1 case.
- Paul's Online Math Notes — Root Test Confirm the analogous root test for terms raised to the nth power.
Sources
- Ratio Test, Root Test, and Absolute Convergence