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Topic 3 of 6

Convergence & Error Bounds

Knowing a series exists is not enough — you need to know when it converges and how large the error is when you truncate it. The Lagrange remainder and ratio test give you precise, quantitative answers to both questions.

Key Concepts

Radius of Convergence
The power series ∑cn(x−a)n\sum c_n(x-a)^n converges absolutely for all ∣x−a∣<R|x-a| < R and diverges for ∣x−a∣>R|x-a| > R. At ∣x−a∣=R|x-a|=R, behavior must be checked separately.
Ratio Test
If L=lim⁡n→∞∣an+1/an∣L = \lim_{n\to\infty}|a_{n+1}/a_n|, the series converges if L<1L<1, diverges if L>1L>1, and the test is inconclusive at L=1L=1.
Lagrange Remainder
The error in the NN-th order Taylor polynomial satisfies ∣RN(x)∣=∣f(N+1)(ξ)(N+1)!(x−a)N+1∣|R_N(x)| = \left|\frac{f^{(N+1)}(\xi)}{(N+1)!}(x-a)^{N+1}\right| for some ξ\xi between xx and aa.
Alternating Series Bound
For an alternating series whose terms decrease in magnitude, the error from stopping after NN terms is bounded by the absolute value of the first omitted term.
Absolute Convergence
If ∑∣an∣\sum |a_n| converges, the series converges absolutely. Absolute convergence implies convergence, and allows rearrangement of terms.

Key Equations

Lagrange Remainder
RN(x)=f(N+1)(ξ)(N+1)! (x−a)N+1R_N(x) = \frac{f^{(N+1)}(\xi)}{(N+1)!}\,(x-a)^{N+1}
For some ξ\xi strictly between xx and aa. Provides an exact (not just asymptotic) error bound.
Radius of Convergence
R=1lim sup⁡n→∞∣cn∣1/nR = \frac{1}{\displaystyle\limsup_{n\to\infty}|c_n|^{1/n}}
The Cauchy–Hadamard formula. More commonly found via the ratio test.
Ratio Test Radius
R=lim⁡n→∞∣cncn+1∣R = \lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|
When the limit exists, this equals the radius of convergence.
Worked Example

Bounding the Error in sin⁡(0.5)\sin(0.5)

Problem

Approximate sin⁡(0.5)\sin(0.5) using 2 terms (x−x3/6x - x^3/6) and find an upper bound on the error.

Solution

The 2-term approximation at x=0.5x=0.5: T3(0.5)=0.5−(0.5)3/6=0.5−0.02083=0.47917T_3(0.5) = 0.5 - (0.5)^3/6 = 0.5 - 0.02083 = 0.47917.

The Lagrange remainder for sin⁡(x)\sin(x) after N=3N=3: the next term is ∣x∣5/5!=∣sin⁡(5)(ξ)∣/5!⋅x5|x|^5/5! = |\sin^{(5)}(\xi)|/5! \cdot x^5. Since ∣sin⁡(5)(ξ)∣≤1|\sin^{(5)}(\xi)| \leq 1:

∣R3∣≤∣x∣55!=(0.5)5120=0.03125120≈2.6×10−4|R_3| \leq \frac{|x|^5}{5!} = \frac{(0.5)^5}{120} = \frac{0.03125}{120} \approx 2.6\times10^{-4}

Alternatively, since sin⁡(x)\sin(x) has an alternating series with decreasing terms at x=0.5x=0.5, the error is bounded by the first omitted term, which is the same bound.

Answer sin⁡(0.5)≈0.47917\sin(0.5) \approx 0.47917 with error ≤2.6×10−4\leq 2.6\times10^{-4} (exact value: 0.479430.47943, error ≈2.6×10−4\approx 2.6\times10^{-4} ✓)
Practice

Exercises

5 problems
1 of 5

Using the alternating series bound, the error in approximating sin⁡(0.1)\sin(0.1) with just one term T1(x)=xT_1(x) = x is bounded by the next term ∣x∣3/3!|x|^3/3!. What is this bound at x=0.1x=0.1? Express as a decimal.

2 of 5

The radius of convergence of ∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n} (the series for −ln⁡(1−x)-\ln(1-x)) is R=?R = ?

3 of 5

Using the alternating series bound, the error in approximating cos⁡(0.5)\cos(0.5) with 2 terms (1−x2/21 - x^2/2) is bounded by the next term ∣x∣4/4!|x|^4/4! at x=0.5x=0.5. What is this bound?

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4 of 5

Evaluate the 4-term partial sum of arctan⁡(x)=x−x3/3+x5/5−x7/7+⋯\arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + \cdots at x=1x=1. This sum is a famous approximation of π\pi. What value do you get?

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5 of 5

The 4-term partial sum of ∑n=1∞(0.5)n/n\sum_{n=1}^{\infty}(0.5)^n/n is S4=0.5+0.52/2+0.53/3+0.54/4S_4 = 0.5 + 0.5^2/2 + 0.5^3/3 + 0.5^4/4. What is S4S_4? (The full series converges to ln⁡2≈0.6931\ln 2 \approx 0.6931.)

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Key Takeaways

  • The Lagrange remainder ∣RN∣≤M∣x−a∣N+1/(N+1)!|R_N| \leq M|x-a|^{N+1}/(N+1)! gives a rigorous upper bound on truncation error.
  • For alternating series with decreasing terms, the error is simply bounded by the first omitted term.
  • The radius of convergence RR satisfies R=lim⁡∣cn/cn+1∣R = \lim|c_n/c_{n+1}| (when the limit exists).
  • For exe^x, sin⁡x\sin x, cos⁡x\cos x: R=∞R = \infty. For ln⁡(1+x)\ln(1+x), binomial (1+x)α(1+x)^\alpha: R=1R = 1.
  • A truncated Taylor series is always more accurate closer to the expansion point. The error scales as (x−a)N+1(x-a)^{N+1}.
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