The Convergence Tests

The Integral Test needs an antiderivative, which most series do not offer. The comparison, alternating, ratio, and root tests need none: each reads convergence off the general term directly, by comparison with a known series, by the sign pattern, or by the ratio of consecutive terms.

The Comparison Test

If a series of positive terms is dominated term by term by a convergent series, its partial sums are bounded, so it converges too. The reverse comparison forces divergence.

The proof is the Monotonic Sequence Theorem again. With $s_{n} = \sum_{i \leq n} a_{i}$ and $t_{n} = \sum_{i \leq n} b_{i}$ , positivity makes both increasing. If $\sum b_{n} = t$ then $s_{n} \leq t_{n} \leq t$ , so ${s_{n}}$ is bounded above and converges. The divergent case is the contrapositive.

Term-by-term comparison: the tested series (dark) sits under a convergent series (light) whose terms cap it at every index, so its partial sums stay bounded.

For comparison the two standard families supply the reference series:

a $p$ -series $\sum 1/ n^{p}$ converges iff $p > 1$ ;
a geometric series $\sum a r^{n - 1}$ converges iff $∣ r ∣ < 1$ .

The direction matters. To conclude convergence the terms must be smaller than a convergent series; being smaller than a divergent one says nothing. The next test removes that restriction.

The Limit Comparison Test

The inequality version fails on series like $\sum 1/ (2^{n} - 1)$ : the natural comparison $1/ (2^{n} - 1) > 1/ 2^{n}$ points the wrong way. Comparing limits of the ratio sidesteps the issue.

If the ratio settles to a positive finite $c$ , the two series are the same size for large $n$ , so they converge or diverge together. For $\sum 1/ (2^{n} - 1)$ , compare with $b_{n} = 1/ 2^{n}$ :

n \to \infty lim \frac{1/ ( 2 ^{n} - 1 )}{1/ 2 ^{n}} = n \to \infty lim \frac{2 ^{n}}{2 ^{n} - 1} = n \to \infty lim \frac{1}{1 - 2 ^{- n}} = 1,

and since $\sum 1/ 2^{n}$ converges, so does the given series. In practice, build $b_{n}$ by keeping only the highest powers of $n$ in numerator and denominator.

Alternating series

A series whose terms alternate in sign,

n = 1 \sum \infty (- 1)^{n - 1} b_{n} = b_{1} - b_{2} + b_{3} - b_{4} + \dots, b_{n} > 0,

converges under a light condition: the magnitudes need only decrease to zero.

The partial sums step right by $b_{1}$ , left by $b_{2}$ , right by $b_{3}$ , and so on. Because the steps shrink, the even partial sums $s_{2}, s_{4}, s_{6}, \dots$ increase while the odd ones $s_{1}, s_{3}, s_{5}, \dots$ decrease, and the two bracket a common limit $s$ from below and above.

Partial sums of an alternating series close in on the sum

s

from both sides: even sums rise, odd sums fall, and each lies within one term of

s

Formally, the even sums are increasing and bounded above by $b_{1}$ , so they converge to some $s$ ; the odd sums converge to the same $s$ because $s_{2 n + 1} = s_{2 n} + b_{2 n + 1}$ and $b_{2 n + 1} \to 0$ . The alternating harmonic series

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{n}

converges by this test (with $b_{n} = 1/ n$ ), even though the harmonic series itself diverges. Its sum is $ln 2$ .

When the monotonicity of $b_{n}$ is not obvious, test the related function's derivative. For $b_{n} = n^{2} / (n^{3} + 1)$ , the function $f (x) = x^{2} / (x^{3} + 1)$ has $f^{'} (x) = x (2 - x^{3}) / (x^{3} + 1)^{2} < 0$ once $x > 32$ , so $b_{n}$ eventually decreases — enough for the test.

Estimating an alternating sum

The bracketing picture also yields an error bound: since $s$ lies between $s_{n}$ and $s_{n + 1}$ , the error in stopping at $s_{n}$ is smaller than the first term left out.

The rule error below the first omitted term is special to alternating series meeting these two conditions; it does not apply to series in general.

Absolute convergence, the Ratio Test, and the Root Test

For series with irregular signs, test the absolute values.

The proof uses $0 \leq a_{n} + ∣ a_{n} ∣ \leq 2∣ a_{n} ∣$ : if $\sum ∣ a_{n} ∣$ converges then $\sum (a_{n} + ∣ a_{n} ∣)$ converges by comparison, and $\sum a_{n} = \sum (a_{n} + ∣ a_{n} ∣) - \sum ∣ a_{n} ∣$ is a difference of convergent series. The alternating harmonic series converges but $\sum 1/ n$ diverges, so it is conditionally convergent. Absolute convergence is the stronger property: it is preserved under rearrangement, whereas a conditionally convergent series can be reordered to sum to any real number (Riemann's rearrangement theorem).

Absolute convergence is a strict subset of convergence. The outer region holds every convergent series; the inner disk holds the absolutely convergent ones; the ring between them holds the conditionally convergent series, such as the alternating harmonic series.

The next two tests detect absolute convergence by measuring how fast $∣ a_{n} ∣$ decays.

The idea is comparison with a geometric series. If $L < 1$ , pick $r$ with $L < r < 1$ ; past some $N$ the ratios stay below $r$ , so $∣ a_{N + k} ∣ \leq ∣ a_{N} ∣ r^{k}$ , and the tail is dominated by a convergent geometric series. If $L > 1$ the terms eventually grow, so $a_{n} \neq \to 0$ and the Test for Divergence applies.

The Ratio-Test limit

L

decides on a number line:

L < 1

forces absolute convergence,

L > 1

forces divergence, and

L = 1

leaves the question open.

The Ratio Test is most effective when $a_{n}$ contains factorials or constants raised to the $n$ th power, where successive terms cancel cleanly. For $\sum (- 1)^{n} n^{3} / 3^{n}$ ,

\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 ) ^{3}}{3 ^{n + 1}} \cdot \frac{3 ^{n}}{n ^{3}} = \frac{1}{3} (1 + \frac{1}{n})^{3} \to \frac{1}{3} < 1,

so the series converges absolutely.

The test is useless on rational functions of $n$ : for every $p$ -series the ratio tends to $1$ .

When the general term is an $n$ th power, the Root Test is cleaner.

For $\sum ((2 n + 3) / (3 n + 2))^{n}$ , taking the $n$ th root gives $(2 n + 3) / (3 n + 2) \to 2/3 < 1$ , so the series converges. If the Ratio Test gives $L = 1$ , the Root Test gives $L = 1$ as well, and conversely; the two are inconclusive on the same series.

A strategy for testing series

There is no fixed order of tests to try; classify the series by the form of its general term and pick the matching test.

Routing a series to a test by the shape of its general term

a_{n}

, from a quick divergence check down to comparison, ratio, root, and integral.

The classification in words:

$a_{n} \neq \to 0$ at a glance: Test for Divergence.
$\sum 1/ n^{p}$ : $p$ -series. $\sum a r^{n - 1}$ : geometric.
$a_{n}$ rational or algebraic in $n$ : comparison or limit comparison with a $p$ -series, choosing $p$ from the highest powers.
Alternating signs: Alternating Series Test.
Factorials or $n$ th powers of a constant: Ratio Test.
$a_{n} = (b_{n})^{n}$ : Root Test.
$a_{n} = f (n)$ with $\int_{1}^{\infty} f$ tractable: Integral Test.

Test	Best for	Key quantity
Divergence	any, quick screen	$lim a_{n}$
Comparison / Limit Comparison	rational, algebraic $a_{n}$	ratio to a $p$ -series
Alternating Series	$(- 1)^{n} b_{n}$	$b_{n} ↓ 0$ ; error $\leq b_{n + 1}$
Ratio	factorials, $c^{n}$	$lim ∣ a_{n + 1} / a_{n} ∣$
Root	$(b_{n})^{n}$	$lim n ∣ a_{n} ∣$