Ratio, root, comparison and alternating series tests

Last week you saw the geometric series, the nth-term test and the integral test. The integral test is powerful but does not work for terms that cannot be integrated (factorials, $\sin n$, complicated fractions). This week we add four tools that you will use far more often. The series part of the exam almost always comes as “Determine whether the series converges or diverges. Name the test you use.” Choosing the right test is half the job; writing the conditions of the test is the other half.

Golden rules:

1. The nth-term test first. If the term does not go to zero, no other test is needed.
2. If you see a factorial or an exponential ($c^n$, $n^n$), use the ratio test. If the whole term is an $n$th power, the root test.
3. If there are only powers, roots, logarithms of $n$, compare (mostly limit comparison) with a p-series.
4. If the sign changes, first look at the absolute value; if it does not converge absolutely, move to the alternating series test.
5. If the ratio or root test gives $L=1$, the test gives no result. You cannot write “converges” or “diverges”.

The ratio test

Theorem. Let $\displaystyle L = \lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$.

• If $L<1$ the series converges absolutely (and therefore converges).
• If $L>1$ or $L=\infty$ it diverges.
• If $L=1$ the test is inconclusive.

Intuition. For large $n$, $a_{n+1}\approx La_n$, so in the tail the series behaves like a geometric series with ratio $L$. A geometric series converged when $\lvert r\rvert<1$.

Example 1. $\displaystyle\sum\frac{2^n}{n!}$.

$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n} = \frac{2}{n+1}\to0<1.$

Converges. In the factorial cancellation we used $\dfrac{n!}{(n+1)!} = \dfrac{1}{n+1}$. Memorize this and the pattern $\dfrac{(2n)!}{(2n+2)!} = \dfrac{1}{(2n+1)(2n+2)}$.

Example 2. $\displaystyle\sum\frac{n^3}{2^n}$. The ratio is $\dfrac{(n+1)^3}{n^3}\cdot\dfrac12\to\dfrac12$. Converges.

PITFALL · $L=1$. $\sum\frac1n$ (divergent) and $\sum\frac{1}{n^2}$ (convergent) both give $L=1$ in the ratio test. So for rational expressions with polynomials or roots, the ratio test can never decide. Do not waste time on these series, move on to comparison.

The root test

Theorem. $L = \displaystyle\lim_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}$. The conclusions are the same as the ratio test: $L<1$ converges absolutely, $L>1$ diverges, $L=1$ no result.

The root test is unrivaled when the term has the form $(\ldots)^n$, because the $n$th root removes the exponent.

Example 3. $\displaystyle\sum\left(\frac{n}{3n+1}\right)^n$. $\sqrt[n]{a_n} = \dfrac{n}{3n+1}\to\dfrac13<1$. Converges.

Helpful limits: $\sqrt[n]{n}\to1$, $\sqrt[n]{c}\to1$. For example, for $\sum\dfrac{n}{2^n}$, $\sqrt[n]{a_n} = \dfrac{\sqrt[n]n}{2}\to\dfrac12$.

Comparison tests

These two tests are used only for positive-term series (at least from some point on).

Direct comparison. If $0\le a_n\le b_n$ for every $n$:

• $\sum b_n$ converges $\Rightarrow$ $\sum a_n$ converges.
• $\sum a_n$ diverges $\Rightarrow$ $\sum b_n$ diverges.

Example 4. $\displaystyle\sum\frac{1}{2^n+5}$. $\dfrac{1}{2^n+5}<\dfrac1{2^n}$ and $\sum\frac1{2^n}$ is a convergent geometric series. Converges.

Limit comparison. If $a_n, b_n>0$ and

$\lim_{n\to\infty}\frac{a_n}{b_n} = c,\qquad 0<c<\infty$

then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. Extreme cases: if $c=0$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $c=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

How do you choose $b_n$? Keep only the highest-degree terms of the numerator and denominator. Drop the constant coefficients.

a_n = (n+1)/(n^3+2)          → dominant: n/n^3 = 1/n^2         → b_n = 1/n^2
a_n = (3n^2 - 1)/sqrt(n^5+n) → dominant: n^2/n^(5/2) = 1/n^(1/2) → b_n = 1/sqrt(n)
a_n = 1/(n - ln n)           → dominant: 1/n                   → b_n = 1/n

Example 5. $\displaystyle\sum\frac{n+1}{n^3+2}$. With $b_n = 1/n^2$: $\dfrac{a_n}{b_n} = \dfrac{n^3+n^2}{n^3+2}\to1$. $\sum1/n^2$ converges, so the series converges.

TIP · why limit comparison? In direct comparison the direction of the inequality is critical and often comes out reversed. $\dfrac{1}{n^2-3}>\dfrac{1}{n^2}$, that is, it is bigger than a convergent series, and direct comparison says nothing. In limit comparison the direction does not matter: the ratio goes to $1$, done.

The alternating series test

Theorem (Leibniz). In the series $\sum(-1)^{n+1}b_n$, if

1. $b_n>0$,
2. $b_n$ is decreasing ($b_{n+1}\le b_n$, from some point on),
3. $b_n\to0$

then the series converges.

Example 6. The alternating harmonic series $1-\frac12+\frac13-\frac14+\dots$ All three conditions hold, so it converges (its sum is $\ln2$). But its absolute value is the harmonic series, which diverges.

Error estimate. If the conditions hold, the difference between $s_n$ and the true sum is smaller than the first term dropped:

$\lvert S-s_n\rvert\le b_{n+1}.$

Example 7. If you compute the sum of $\sum\dfrac{(-1)^{n+1}}{n^2}$ with $s_4$, the error is at most $b_5 = \dfrac{1}{25} = 0.04$.

Absolute and conditional convergence

Theorem. Every absolutely convergent series is convergent. That is why, for series with irregular signs (those containing $\sin n$, $\cos n$), you switch to the absolute value and compare.

The template for an “absolutely, conditionally, or diverges?” question in the exam:

1) lim a_n ≠ 0 ? → diverges (done)
2) Does Σ|a_n| converge? (p-series, comparison, ratio) → if yes, ABSOLUTELY
3) If not: alternating series test (b_n decreasing, → 0) → if it holds, CONDITIONALLY

Which test? The decision table

Mixed examples: choosing the test

Example 8. $\displaystyle\sum\frac{3^n}{n^2\,4^n}$. There is an exponential term. Ratio: $\dfrac34\cdot\dfrac{n^2}{(n+1)^2}\to\dfrac34<1$. Converges. (Direct comparison would also work: $\le(3/4)^n$.)

Example 9. $\displaystyle\sum\frac{\sqrt n}{n^2+1}$. Only powers. Dominant: $\dfrac{n^{1/2}}{n^2} = \dfrac{1}{n^{3/2}}$. The limit comparison ratio $\to1$, $p = \frac32>1$, converges.

Example 10. $\displaystyle\sum\frac{(-1)^n n}{n^2+1}$. The absolute value is $\approx\frac1n$, diverges. $b_n = \dfrac{n}{n^2+1}\to0$ and for $x\ge1$, $f'(x) = \dfrac{1-x^2}{(x^2+1)^2}\le0$, decreasing. The alternating test holds: conditionally convergent.

Example 11. $\displaystyle\sum\left(1-\frac1n\right)^n$. It looks like a root test case but be careful: the term itself $\to e^{-1}\neq0$. It diverges by the nth-term test. The root test here gives $\sqrt[n]{a_n} = 1-\frac1n\to1$ and would be silent; if you had skipped the first step you would have been stuck.

Common mistakes