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Series convergence tests: which test to use, and what each one can't tell you

An infinite series a₁ + a₂ + a₃ + ⋯ converges when its partial sums settle on a finite number. A handful of tests settle most textbook series, and the work goes wrong in two ways: a test is used outside the conditions that make it work, or its answer is read in a case where it gives none. Knowing which test fits the shape of the terms, and what each one can't say, is most of the job.

Why it works

Every test compares the series with something whose behavior is already known. A geometric series has partial sums with a closed form, a(1 − r^N)/(1 − r), and r^N → 0 exactly when |r| < 1. The integral test treats the terms as rectangles of width 1 under or over the graph of f(x), so the sum and the integral are finite or infinite together. That's where the p-series rule comes from: ∫₁^∞ x^−p dx is finite only for p > 1.

The comparison tests pass that knowledge along. A positive series smaller than a convergent one has partial sums that are bounded and increasing, so it converges; a series bigger than a divergent one is pushed to infinity. The ratio test compares with a geometric series: if each term is eventually about ℓ times the one before, the series behaves like ∑ ℓⁿ. At ℓ = 1 that comparison says nothing, because ∑ 1/n and ∑ 1/n² both have ratio limit 1, and one diverges while the other converges.

Example 1: a geometric series that starts at n = 2

Find ∑_n=2^∞ 34ⁿ.

1. r = 14, |r| < 1Each term is a quarter of the one before, so the series converges.
2. first term = 34² = 316The sum starts at n = 2, so that's the first term in it.
3. S = 3/161 − 1/4 = 3/163/4 = 14First term over 1 − r. Check: 3/16 + 3/64 + 3/256 = 0.246…, climbing toward 0.25.

Example 2: limit comparison for a ratio of polynomials

Does ∑_n=1^∞ n + 5n² + 1 converge?

1. n + 5n² + 1 behaves like nn² = 1nFor large n, only the highest power on top and on the bottom matters.
2. (n + 5)/(n² + 1)1/n = n² + 5nn² + 1 → 1Limit comparison with b_n = 1/n. The limit is positive and finite.
3. ∑ 1n diverges, so ∑ n + 5n² + 1 divergesThe harmonic series is a p-series with p = 1. The terms do go to 0, so the divergence test couldn't have settled this one.

Example 3: the integral test, with bounds on the sum

Show that ∑_n=1^∞ 1n² + 1 converges, and bound its sum.

1. f(x) = 1x² + 1 is positive and decreasing for x ≥ 1The conditions of the integral test. Check them before integrating.
2. ∫₁^∞ dxx² + 1 = lim_b→∞ (arctan b − arctan 1) = π2 − π4 = π4An improper integral with a finite value, so the series converges.
3. π4 ≤ S ≤ 12 + π4, so 0.785 ≤ S ≤ 1.285The integral from 1 is below the sum, and the first term plus the integral is above it. The sum is about 1.0767. The integral's value, π/4, is not the sum.

Example 4: the ratio test

Does ∑_n=1^∞ n²2ⁿ converge?

1. a_n+1a_n = (n + 1)²2ⁿ⁺¹ · 2ⁿn² = 12(n + 1n)²The 2ⁿ cancels, leaving one factor of 1/2.
2. ℓ = lim 12(1 + 1n)² = 12The polynomial part of the ratio tends to 1, so the exponential decides.
3. ℓ = 12 < 1, so the series convergesIts sum happens to be exactly 6. Any polynomial over an exponential with base above 1 converges the same way.

Example 5: an alternating series, and how many terms you need

Show that ∑_n=1^∞ (−1)ⁿ⁺¹√n converges, decide whether it converges absolutely, and find how many terms guarantee an error below 0.1.

1. b_n = 1√n decreases to 0, so the series convergesThe alternating series test needs both: the terms shrink, and they shrink to 0.
2. ∑ 1√n divergesA p-series with p = 1/2. So the convergence is conditional, not absolute.
3. error after N terms ≤ b_N+1 = 1√(N + 1) < 0.1 needs N + 1 > 100The error is at most the first term left out.
4. N = 100 termsWith 99 terms the bound is exactly 0.1, not below it. The sum is about 0.6049, and S₁₀₀ ≈ 0.5550 is off by about 0.050, inside the bound.

Common mistakes, and the exact line they happen on

1. Starting a geometric sum at the wrong term

The series from Example 1.

1. ∑_n=2^∞ 3 (14)ⁿ, r = 14Right so far.
2. S = 31 − 1/4 = 4This is the mistake. The 3 is the term at n = 0, and the sum starts at n = 2. The formula a/(1 − r) assumes the sum starts at r⁰; in general it is first term over 1 − r. This answer includes 3 + 3/4 that aren't in the series: 4 − 3 − 3/4 = 1/4, the right sum.

2. Reading terms that go to 0 as convergence

The harmonic series.

1. 1n → 0Right so far.
2. so ∑ 1n convergesThis is the mistake. The divergence test runs one way only: terms that don't go to 0 force divergence, but terms that do go to 0 prove nothing. The harmonic series diverges, slowly: S₁₀₀₀ ≈ 7.49 and S_1,000,000 ≈ 14.39, growing like ln N without bound.

3. Comparing in the direction that proves nothing

The series ∑ n + 1n².

1. n + 1n² ≥ 1n², and ∑ 1n² convergesBoth facts are true.
2. so ∑ n + 1n² convergesThis is the mistake. Being bigger than a convergent series proves nothing: a series of 1s is bigger too. To prove convergence, the terms must be at most those of a convergent series; to prove divergence, at least those of a divergent one. Here (n + 1)/n² ≥ 1/n, and the series diverges.

Practice

1. Find ∑_n=0^∞ (−12)ⁿ.

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   Geometric with first term 1 and r = −12, so the sum is 11 + 1/2 = 23.

2. Find ∑_n=1^∞ 1n(n + 2).

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   1n(n + 2) = 12(1n − 1n + 2), and the sum telescopes, leaving 12(1 + 12) = 34.

3. Does ∑_n=1^∞ 3ⁿn! converge?

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   Yes. The ratio is 3n + 1 → 0 < 1. The sum is e³ − 1 ≈ 19.09.

4. Does ∑_n=2^∞ (−1)ⁿln n converge absolutely, conditionally, or not at all?

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   Conditionally. 1/ln n decreases to 0, so the alternating series test applies, but 1/ln n ≥ 1/n and ∑ 1/n diverges, so ∑ 1/ln n diverges.

5. Find ∑_n=1^∞ 2ⁿ + 13ⁿ.

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   Split it into two geometric series: ∑ (23)ⁿ + ∑ (13)ⁿ = 2 + 12 = 52, each starting at n = 1.