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10.10 Alternating Series Error Bound

Keywords

Alternating Series Error Bound 交错级数误差项 • Remainder (Error) 余项/误差

Partial Sum 部分和 • Truncation Error 截断误差

Estimating the Sum

When we calculate a limit or a sum, we often use a partial sum (\(S_n\)) as an approximation. The difference between the actual infinite sum (\(S\)) and our approximation is called the Remainder (\(R_n\)), or simply the Error.

\[ \text{Error} = |R_n| = |S - S_n| \]

The Error Bound Theorem

For alternating series, the error is always "trapped" by the next term in the sequence.

核心要点：Alternating Series Error Bound (LIM-7.B.1)

If an alternating series \(\sum (-1)^n a_n\) satisfies the conditions of the Alternating Series Test (terms are decreasing in magnitude and \(\lim_{n \to \infty} a_n = 0\)), then:

The error in using the \(n\)-th partial sum \(S_n\) to approximate the total sum \(S\) is less than or equal to the magnitude of the first omitted term (\(a_{n+1}\)).

\[ |S - S_n| \le a_{n+1} \]

Visualizing the Error

Think of the partial sums as a sequence of jumps. Because each jump is smaller than the previous one and in the opposite direction, the actual sum \(S\) is always "trapped" between \(S_n\) and \(S_{n+1}\).

In the graph above, notice that the distance from any point (partial sum) to the red line (actual sum) is always smaller than the vertical distance to the next point.

Example: Finding the Error

Approximate the sum of \(\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\) using the first 3 terms and find the error bound.

  1. Partial Sum \(S_3\):

    \[ S_3 = 1 - \frac{1}{4} + \frac{1}{9} = \frac{36 - 9 + 4}{36} = \frac{31}{36} \approx 0.8611 \]

  2. First Omitted Term (\(a_4\)): The next term in the series (ignoring the sign) is \(a_4 = \frac{1}{4^2} = \frac{1}{16} = 0.0625\).

  3. Conclusion: The actual sum \(S\) is within \(0.0625\) of \(0.8611\). Mathematically: \(0.7986 \le S \le 0.9236\).

Extension: Error vs. Accuracy

In computer science and numerical analysis, this theorem is vital. If you need a result accurate to \(10^{-6}\) (six decimal places), you simply find the smallest \(n\) such that \(a_{n+1} < 0.000001\). This tells you exactly how much work (how many iterations) the computer needs to do to guarantee the result is correct.

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