Alternating Series Error Bound
Introduced| English |
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| alternating series error bound/ˈɔːltəneɪtɪŋ ˈsɪəriːz ˈerə baʊnd/ |
How close is a partial sum?
- If you stop an alternating series after a few terms, how far off is your estimate?
- For a convergent alternating series, there's a beautifully simple bound.
- The alternating series error bound 交错级数误差界: the error is no bigger than the first omitted term.
- One term tells you the accuracy — no messy computation.
The bound
- Approximate the sum $S$ by the $n$th partial sum $S_n$. The remainder (error) is $|S-S_n|$.
- For a convergent alternating series (decreasing $b_n\to0$):
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$$|S-S_n|\le b_{n+1}$$
- The error is at most the absolute value of the next term — the first one you left out.
The next term bounds the error
Partial sums of an alternating series trap the true sum, so the error is at most the size of the next (omitted) term.
For a convergent alternating series, the error $|S-S_n|$ is at most...
$|S-S_n|\le b_{n+1}$.
The error is bounded by the first ____ term, not the last included one.
Use $b_{n+1}$, the term you did not add.
Why the next term bounds it
- Because terms alternate and shrink, the true sum is always trapped between consecutive partial sums.
- So stopping at $S_n$ overshoots or undershoots by less than the size of the next term.
- The partial sums close in like a shrinking accordion around $S$.
- Hence the very next term is a guaranteed error ceiling.
Approximating $1-\tfrac12+\tfrac13-\cdots$ with $3$ terms, the error bound is the next term $b_4$. What is it?
$b_4=\tfrac14=0.25$.
The true sum is always trapped between consecutive partial sums of an alternating series.
That is why the next term bounds the error.
Using it to guarantee accuracy
- Want the error under $0.01$? Find the first term with $b_{n+1}\le 0.01$ and stop there.
- The bound lets you decide how many terms you need for a target accuracy.
- It only applies to alternating series meeting the Alternating Series Test conditions.
- Simple, tight, and exam-friendly.
This error bound applies to any convergent series, alternating or not.
Only alternating series meeting the test conditions.
To guarantee an error under $0.01$, you keep terms until the next term $b_{n+1}$ is...
Stop when $b_{n+1}\le0.01$.
This bound works only for alternating series that satisfy the Alternating Series Test (decreasing $b_n\to0$) — it does not apply to positive-term series. The error is bounded by the first omitted term $b_{n+1}$, not the last included one. Use $b_{n+1}$, the size of the term you didn't add.
Estimate $\sum_{n=1}^\infty\tfrac{(-1)^{n+1}}{n}=1-\tfrac12+\tfrac13-\cdots$ with the first $3$ terms; bound the error.
- $S_3=1-\tfrac12+\tfrac13=\tfrac{5}{6}\approx0.833$.
- First omitted term: $b_4=\tfrac14$. So $|S-S_3|\le\tfrac14=0.25$.
- The true sum $\ln 2\approx0.693$ is indeed within $0.25$ of $0.833$. ✓
The alternating series error bound: for a convergent alternating series, the error of the $n$th partial sum is at most the first omitted term, $|S-S_n|\le b_{n+1}$. It applies only to alternating series meeting the test's conditions, and lets you pick how many terms guarantee a target accuracy.