Maclaurin Series Table for Quick Reference

A common mistake is to treat a Maclaurin series table as a list of formulas to memorise. That misses the point. The table is a starting point. Every series in it comes from a single foundation: the geometric series, the binomial theorem, or a Taylor series centred at zero. The value of the table is not the formulas themselves, but the fact that you can build almost any other series you need by starting from these and applying substitution, differentiation, or integration. The standard Maclaurin series table lists each series, its convergence interval, and how to apply the list without blindly copying it.

Each series in the table comes from a single operation applied to a known base series.

• Geometric series: 1/(1-x) = sum_{n=0}^{∞} x^n. This is the definition of the sum of a geometric sequence with ratio x; it converges to the closed form when |x| < 1.
• e^x = sum_{n=0}^{∞} x^n/n!. This is the Taylor series of exponential function centred at 0, using f^{(n)}(0) = 1 for all n.
• sin x = sum_{n=0}^{∞} (-1)^n x^{2n+1}/(2n+1)!. Taylor series of sine at 0, using the pattern of derivatives: sin(0)=0, cos(0)=1, -sin(0)=0, -cos(0)=-1, and so on.
• cos x = sum_{n=0}^{∞} (-1)^n x^{2n}/(2n)!. Taylor series of cosine at 0, using the same derivative pattern offset by one.
• ln(1+x) = sum_{n=1}^{∞} (-1)^{n-1} x^n / n. Integrate the geometric series for 1/(1+x) term by term.
• arctan x = sum_{n=0}^{∞} (-1)^n x^{2n+1}/(2n+1). Integrate the geometric series for 1/(1+x^2) term by term.

The six series above are the building blocks. You can create a new power series representation for a function not in the table by substituting, differentiating, or integrating one of these within its interval of convergence. For example, to get the series for e^{-x^2}, substitute -x^2 for x in the e^x series: sum_{n=0}^{∞} (-1)^n x^{2n}/n!. This converges for all x because the original series for e^x converges for all x. To get the series for x sin(x^2), substitute x^2 into the sine series, then multiply every term by x: sum_{n=0}^{∞} (-1)^n x^{4n+3}/(2n+1)!, also convergent for all x...(k-n+1)/n! and (k choose 0)=1. It converges for |x| < 1, and at the endpoints the behaviour depends on k. […] These are the same six in the table above. For the exam you must also be able to state the interval of convergence for each one. The ratio test gives the radius for each series, and you must test endpoints separately. […] A common exam mistake is to state the interval as |x| < 1 without checking endpoints. The College Board awards credit only for the correct interval, including correct endpoint notation.

A list of power series is only useful if you know what to do with it. Here are the three most common operations, each with a concrete example.

• Substitution. To get the series for e^{2x}, replace every x in the e^x series with 2x: sum_{n=0}^{∞} (2x)^n/n! = sum_{n=0}^{∞} 2^n x^n / n!. […] Start from n=1. […] The radius stays 1, but the interval may change. Check endpoints.

These operations are valid inside the radius of convergence. […] "

For the Maclaurin series of e^x, the ratio test gives limit |a_{n+1}/a_n| = |x|/(n+1), which tends to 0 for any x. Since 0 < 1, the series converges absolutely for every x, so the radius is ∞. This means you can evaluate the series at x = 10 or x = -100 and get the correct value, but the number of terms needed for a given accuracy grows with |x|. For hand calculation, use the alternating series estimation theorem to bound the error: the error after truncating at the nth term is less than the (n+1)th term. […] This makes it ideal for error estimation using the alternating series estimation theorem: the absolute error after truncating at the term of degree 2n+1 is less than the first omitted term, which is |x|^{2n+3}/(2n+3)!. For small x, this bound is tiny even with one term. For sin(0.1), using just the first term x = 0.1 gives an error less than 0.1^3/6 ≈ 0.000167. The series converges for all x, but for |x| > 1 you need more terms to keep the error small.

The binomial series expands (1+x)^k for any real k. The series is sum_{n=0}^{∞} (k choose n) x^n, where (k choose n) = k(k-1)...(k-n+1)/n!. […] The endpoints must be tested separately and depend on k. […] For k = 1/2, the series for sqrt(1+x) converges at x = 1 but not at x = -1. The binomial series is the tool for series like 1/(1+x^2)^0.3 or cube root of (1+x).

Copy the table above into a document set to landscape orientation to create a printable version.

[…] The table fits on one side of a single sheet if you keep the expanded form to three terms. […] This gives you a complete reference for a closed-book exam or a quick check during problem solving.

The single thing that most often goes wrong with a Maclaurin series table is that students treat the interval of convergence as a separate fact to memorise instead of a consequence of the ratio test and endpoint testing. If you know how to find the radius from the coefficients and can apply the alternating series test or p-series test to the endpoints, you do not need to memorise any interval. The table is a starting point for deriving series, not an answer key for convergence questions. Outside the radius, the series is not equal to the function. Inside the radius, the series and the function are identical only in the limit as the number of terms goes to infinity. The partial sum is an approximation, not the truth.