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Reference · Calculus

Series & Convergence

A copy-able sheet — Maclaurin expansions with their intervals, every convergence test, and the closed-form sums.

Showing 30 of 30. Click any formula to copy it.

C(k, n) is the generalised binomial coefficient k(k − 1)…(k − n + 1) ÷ n!, which is why the binomial series does not terminate unless k is a non-negative integer. Every interval of convergence above is stated for real x.

A Maclaurin series rewrites a function as an infinite polynomial. The five worth memorising are eˣ, sin x, cos x, ln(1 + x), and 1 ÷ (1 − x). The first three converge for every x; the last two only on an interval — 1 ÷ (1 − x) = 1 + x + x² + … needs |x| < 1.

Why a function becomes a polynomial

A Taylor series is built from derivatives at a single point: each term matches one more derivative of the original function there, so the partial sums hug the curve ever more tightly near that point. A Maclaurin series is the same construction centred at zero, which is why it is the version tabulated — most standard functions are cleanest there.

The catch is that matching derivatives at one point does not guarantee agreement everywhere. The interval of convergence says how far the polynomial can be trusted, and it is as much a part of the formula as the terms themselves. Writing 1 ÷ (1 − x) = Σxⁿ without the condition |x| < 1 gives nonsense at x = 2, where the left side is −1 and the right side diverges.

Derive rather than memorise

Most expansions on the sheet follow from a handful by substitution. Replacing x with −x² in the geometric series gives the series for 1 ÷ (1 + x²), and integrating that term by term gives the arctangent series. Differentiating the sine series gives the cosine series. Four or five memorised expansions plus these moves cover nearly everything a course asks for.

The Maclaurin series worth knowing

Every interval below is stated for real x. Outside it the series says nothing about the function.

FunctionSeriesConverges for
eˣ1 + x + x²/2! + x³/3! + …All x
sin xx − x³/3! + x⁵/5! − …All x
cos x1 − x²/2! + x⁴/4! − …All x
1 ÷ (1 − x)1 + x + x² + x³ + …|x| < 1
ln(1 + x)x − x²/2 + x³/3 − …−1 < x ≤ 1
arctan xx − x³/3 + x⁵/5 − …|x| ≤ 1
(1 + x)ᵏ1 + kx + k(k − 1)x²/2! + …|x| < 1

Choosing a convergence test

Work down the list — the cheap tests rule out most series before the expensive ones are needed.

If the series has…UseNote
Terms that do not approach 0nth-term testProves divergence only, never convergence
A constant ratio between termsGeometric testConverges when |r| < 1, to a ÷ (1 − r)
The form 1 ÷ nᵖp-series testConverges when p > 1; p = 1 is the harmonic series
Factorials or nth powersRatio testInconclusive when the limit equals 1
n appearing as an exponentRoot testOften easier than the ratio test here
A term you can integrateIntegral testNeeds positive, decreasing, continuous terms
Alternating signsAlternating series testTerms must decrease monotonically to 0
A resemblance to a known seriesComparison or limit comparisonCompare against a p-series or geometric series

The traps in convergence questions

The nth-term test is the one most often misused. If the terms approach zero, the test says nothing at all — the harmonic series Σ1 ÷ n has terms going to zero and still diverges. The test can only ever prove divergence.

Two more are worth naming. Absolute and conditional convergence are different things: Σ(−1)ⁿ ÷ n converges, but Σ1 ÷ n does not, so that series is conditionally convergent and its terms can be rearranged to sum to any value you like. And the endpoints of an interval of convergence have to be tested separately — the ratio test gives the radius but is always inconclusive exactly at the ends, which is why ln(1 + x) converges at x = 1 but not at x = −1.

What is the difference between a Taylor and a Maclaurin series?
A Maclaurin series is simply a Taylor series centred at zero. Taylor’s general form expands about any point a using f⁽ⁿ⁾(a)(x − a)ⁿ ÷ n!, and setting a = 0 gives the Maclaurin version that reference tables list.
How do I find the interval of convergence?
Apply the ratio test to get the radius, then test each endpoint separately by substituting it in and using another test. The ratio test is always inconclusive at the endpoints, which is why they need their own check.
Which convergence test should I reach for first?
Check whether the terms approach zero — if they do not, the series diverges immediately. Then look for a geometric or p-series shape. Factorials and nth powers point to the ratio test, and alternating signs to the alternating series test.
Why does the harmonic series diverge if its terms go to zero?
Terms approaching zero is necessary for convergence but not sufficient. The harmonic terms 1 ÷ n shrink too slowly, and the integral test against ∫dx ÷ x = ln x — which itself grows without bound — shows the sum does too.
What is conditional convergence?
A series converges conditionally when it converges but the series of absolute values does not. The alternating harmonic series is the standard example: its sum is ln 2, yet its terms can be rearranged to converge to any number at all.
Do I need to memorise every series in the table?
No. Memorise eˣ, sin x, cos x, and the geometric series, then derive the rest. Substituting −x² into the geometric series and integrating gives arctan x; differentiating sine gives cosine; integrating the geometric series gives ln(1 + x).
How accurate is a truncated series?
For an alternating series whose terms decrease to zero, the error is no larger than the first omitted term. For the general case use the Lagrange remainder, which bounds the error by M|x − a|ⁿ⁺¹ ÷ (n + 1)! where M bounds the next derivative.