Partial Sums and Approximation Error

You have an infinite series, and you need a usable number from it, not a philosophical limit. The single skill that turns an infinite representation into a computable finite sum is the partial sum, and the single skill that keeps it honest is the approximation error bound. The exact question is how the infinite becomes finite and how you verify the result is accurate enough for your work.

A partial sum S_N(x) truncates the series after N+1 terms. For a power series centered at a, S_N(x) = sum_{n=0}^N c_n (x-a)^n. That is what you actually evaluate when you type a value into any calculator, the series sum calculator evaluates this truncated version, not the infinite series itself. The difference between the true function value and S_N(x) is the remainder R_N(x) = f(x) - S_N(x). That remainder is the error you need to bound.

Every claim here is recomputed from standard expansions. Work one example end to end: approximate sin(0.5) using its Maclaurin series. The series is sin x = x - x^3/3! + x^5/5! - ... . With one term (N=1), S_1(0.5) = 0.5. The actual sin(0.5) is 0.4794. Your error is 0.0206. That number came from a real function, not a guess.

The failure case occurs when the series is evaluated outside its radius of convergence. Use the same series at x=2. The Maclaurin series for sin x converges for all real x, that works, but try the geometric series 1/(1-x) = sum x^n at x=1.2. The series diverges. S_5(1.2) = 1 + 1.2 + 1.44 + 1.728 + 2.0736 + 2.4883 = 10.9435. The true value is 1/(1-1.2) = -5. A partial sum outside the radius does not approach the function; it lies.

To compute the error for a given partial sum, you need a bound, not the exact remainder, you do not know f(x) exactly, or you would not be approximating. Two standard tools cover nearly every case: the Lagrange error bound and the alternating series estimation theorem.

The Lagrange bound applies to any Taylor series where the function is sufficiently differentiable. It says |R_N(x)| less than or equal to M times |x-a|^(N+1) divided by (N+1)!. M is the maximum absolute value of the (N+1)th derivative on the closed interval between the center a and the evaluation point x.

Apply it to sin(0.5). The third derivative of sin x is -cos x. On the interval from 0 to 0.5, the maximum of | -cos t | is 1 (cosine peaks at 0). So |R_1(0.5)| less than or equal to 1 times (0.5)^3 divided by 6. That is 0.125 divided by 6 equals 0.02083. The actual error 0.0206 is below this bound. The bound works.

For alternating series like sin x or cos x, the alternating series estimation theorem gives a simpler bound: |R_N| is less than or equal to the absolute value of the first omitted term. For sin(0.5) with one term, the first omitted term is -(0.5)^3/6 = -0.02083. Its absolute value is 0.02083, the same Lagrange bound in this case. The theorem requires the series to be alternating, with terms decreasing in absolute value and approaching zero.

The explanation of the calculation connects directly to the question: how do you know the partial sum is close enough? You do not guess. You compute a bound and compare it to your tolerance.

The Ratio Test Finds the Radius

Before you can trust any partial sum, you must know the radius of convergence R. The ratio test gives it: R = limit of |a_n / a_{n+1}| as n approaches infinity. For sin x, the coefficients are zero for even n and alternating for odd n. The ratio of successive nonzero terms gives R = infinity, the series converges for all x. For the geometric series 1/(1-x), the coefficients are all 1, so the limit is 1. The series converges for |x| less than 1.

Interval of Convergence Includes Endpoint Checks

R gives the width, but the interval of convergence requires testing the endpoints x = a plus or minus R separately. For the geometric series centered at 0, R=1, so check x=1 and x=-1. At x=1, the series is sum 1^n, which diverges. At x=-1, the series is sum (-1)^n, which diverges because the terms do not approach zero. The interval is open (-1, 1). Many students miss this and get the interval wrong.

One Concrete Failure Mode

The Lagrange bound fails when the function is not (N+1)-times differentiable on the interval. Take f(x) = |x| at center a=0. The first derivative does not exist at 0. You cannot compute a Lagrange bound for any N. The series representation is not a Taylor series because the function is not smooth. If you need to approximate |x| near 0, use a different method, like a piecewise linear approximation or a rational function.

The Lagrange bound is an upper bound, not the exact error. It may overestimate the true remainder if you overestimate M. In practice, you find M by maximizing the derivative on the interval, which can be difficult for complicated functions. If you pick M too small, the bound is invalid and your approximation may be less accurate than you claim.

The alternating series bound is tighter and easier to compute, but it applies only to series that meet its conditions. Check that terms alternate in sign, decrease monotonically in absolute value, and approach zero. If the series is not alternating, do not use this bound.

The most common mistake is to treat the partial sum as the function value without stating the error. If you report S_N as the value of f(x) at a point, you must also report the bound on the remainder. Otherwise, you have misled the reader. This is not pedantry; it is the definition of approximation.