Alternating series estimation theorem and error bound

The alternating series test can tell us that a series converges, but we may still want to know what it adds up to. If we stop after a few terms, the size of the next term tells us how far off our estimate can be. Let’s use that bound to decide when we have enough terms.

Use the first omitted term to bound the error

Write the alternating terms as (-1)n-1bn with bn>0. If bn+1≤bn and bn→0, the alternating series converges. Let S be its sum and Sn be the sum through term n. The remainder is Rn=S-Sn.

|Rn|=|S-Sn|≤bn+1

The consecutive partial sums Sn and Sn+1 sit on opposite sides of S, and their difference in size is bn+1. That’s why the distance from Sn to S is at most the size of the first term we left out.

Round a series estimate to three decimal places

S=∑n=1∞(-1)n-1n10n

Here bn=n/10n. For n≥1, the ratio bn+1/bn=(n+1)/(10n) is at most 1/5. The magnitudes decrease and tend to zero, so the theorem applies.

The first four partial sums
TermsPartial sum
10.1000
20.0800
30.0830
40.0826

The error after three terms satisfies |R3|≤b4=4/104=0.0004. Also, S lies between S4=0.0826 and S3=0.0830. Every number between those endpoints rounds to 0.083 at three decimal places.

Choose the number of terms for a target error

For the alternating harmonic series, bn=1/n decreases to zero. To guarantee an error of at most 0.001, require the first omitted term to be at most that large.

1n+1≤0.001n+1≥1000n≥999

The theorem guarantees the requested accuracy after 999 terms. If the magnitudes instead are bn=1/n!, they also decrease to zero, but much faster. In that case we need (n+1)!≥1000. Since 6!=720 and 7!=5040, the bound is small enough after just six terms.

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