Alternating Series Error Bound
Introduced| English | Meaning |
|---|---|
| alternating series error bound/ˈɔːltəneɪtɪŋ ˈsɪəriːz ˈerə baʊnd/ | alternating series error bound |
| first omitted term | first omitted term |
| remainder/rɪˈmeɪndə/ | remainder |
How accurate is a finite estimate?
- An alternating adjustment model may need only a finite number of terms. The question is how far its partial sum could be from the infinite total.
- The alternating series error bound 交错级数误差界 gives a guaranteed upper bound on the error. It does not usually give the actual error or require knowing the exact infinite sum.
Name the first term you did not include
- For a series starting at $n=1$, let $S_N$ be the sum of its first $N$ terms. The first omitted term 首个未取项 has index $N+1$, and its positive magnitude is $b_{N+1}$.
- For a series starting at index $m$, the first $N$ terms end at $K=m+N-1$. The first omitted index is $K+1=m+N$; a term count and an index need not be the same.
The bound uses the first ____ term, not the last included one.
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The next term has not been added to the approximation.
Check the whole omitted tail
- The omitted tail must alternate with positive, non-increasing magnitudes tending to zero. Under these conditions, the remainder 余项 is $R_N=S-S_N$, and its magnitude satisfies $|S-S_N|\le b_{N+1}$ for a series starting at $1$.
- Convergence alone is not enough to justify this bound. A convergent alternating series with non-monotone magnitudes needs a separate error argument unless a valid monotone tail has been established.
When the alternating-series bound applies, $|S-S_N|$ is at most...
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Use the positive magnitude of the first omitted term.
Which properties are required by this error-bound argument on the omitted tail? Select all.
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Check the positive magnitudes throughout the omitted alternating tail.
The signed remainder R_N=S−S_N must always be non-negative.
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The error is its absolute value; the signed remainder can have either sign.
This first-omitted-term theorem guarantees the same bound for every convergent positive-term series.
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Its alternating omitted-tail hypotheses are essential; positive series need another error argument.
Enclose the sum between consecutive totals
- The exact sum lies between $S_N$ and $S_{N+1}$ under the omitted-tail conditions. These endpoints differ by $|a_{N+1}|=b_{N+1}$.
- If the first omitted term is positive, $S_N$ is a lower estimate; if it is negative, $S_N$ is an upper estimate. Use the sign rather than memorising odd/even rules without checking the first sign.
Under the omitted-tail conditions, the exact sum lies between two consecutive exact partial sums.
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The first omitted magnitude is the width of this enclosure.
Work the three-term harmonic estimate exactly
- For $1-1/2+1/3-\cdots$, $S_3=5/6$. The first omitted term is $-1/4$, so the error is at most $1/4$ and $S_4=7/12$.
- Therefore $7/12\le S\le5/6$, with $S_3$ an upper estimate. The value $1/4$ is an error bound, not a statement that the actual error equals $1/4$.
An enclosure from consecutive partial sums
Default harmonic S_3=5/6 and S_4=7/12 enclose the unknown sum. Compare the exact next-term certificate, outward-rounded decimal endpoints, next sign and shifted denominators.
For the three-term positive-first harmonic estimate, what is the first omitted magnitude?
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The first omitted index is 4, so its magnitude is 1/4.
Find $S_3=1-1/2+1/3$.
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The exact value is 5/6, before rounding.
For that positive-first harmonic series, what is the signed first omitted term after three terms?
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Its sign is negative; its magnitude is 1/4.
The first omitted harmonic term is negative. Therefore the exact S_3 is...
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The remainder points downward from S_3 toward the true sum.
If the next-term bound is 1/4, the actual error must equal 1/4.
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An upper bound is not usually the actual error.
Match the certificate to the requested inequality
- To certify error at most $\varepsilon$ using $|S-S_N|\le b_{N+1}$, it is sufficient to require $b_{N+1}\le\varepsilon$. To certify error strictly below $\varepsilon$ using this non-strict bound alone, require $b_{N+1}<\varepsilon$.
- For harmonic magnitudes and target $0.01$, these conditions give $N\ge99$ for “at most” and $N\ge100$ for the strict certificate. The omitted indices are $100$ and $101$: stop before the omitted term. These are sufficient certificate counts, not claims about the smallest possible actual error.
For harmonic magnitudes, find the smallest integer N satisfying $1/(N+1)\le0.01$.
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The first omitted index must be at least 100, so N≥99.
Using the strict next-term condition $1/(N+1)<0.01$, find the smallest integer N.
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The first omitted index must exceed 100, so N≥100. This is a sufficient certificate count.
Using only $|S-S_N|\le b_{N+1}$, which condition certifies $|S-S_N|<0.01$?
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A strict next-term bound is sufficient for a strict certificate; a non-strict bound alone yields only a non-strict conclusion.
Keep a coefficient and power in the next magnitude
- For $\sum_{n=1}^{\infty}(-1)^n3/n^2$, the bound after $N$ terms is $3/(N+1)^2$. To make this at most $0.001$, require $(N+1)^2\ge3000$.
- The smallest integer satisfying this certificate is $N=54$: $3/55^2\le0.001$, whereas $3/54^2>0.001$. Use the next index and keep the coefficient $3$.
Find the smallest integer N satisfying $3/(N+1)^2\le0.001$.
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N+1 must be at least 55; 54 included terms meet this certificate.
A shifted starting index changes the omitted term
- For $\sum_{n=3}^{\infty}(-1)^{n-3}/n^2$, four included terms have indices $3,4,5,6$. Their sum is $73/1200$, and the first omitted magnitude is $1/49$ at index $7$.
- The next term is positive, so $73/1200\le S\le73/1200+1/49$. Substituting $N+1=5$ into the original denominator would confuse the four-term count with the last index.
A series starts at index 3. What is the last included index after four terms?
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The included indices are 3,4,5,6.
For magnitudes 1/n² starting at index 3, what is the omitted magnitude after four terms?
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The next index is 7, giving 1/7²=1/49.
Allow for rounding the estimate too
- The theorem bounds error from the exact partial sum. If a rounded value $r$ is reported, use $|S-r|\le|S-S_N|+|S_N-r|$ and add the rounding allowance.
- Rounding $S_3=5/6$ to $0.833$ adds $1/3000$ to its $1/4$ bound. Rounding an enclosure must be outward: for example, $[0.583333,0.833334]$ safely contains $[7/12,5/6]$.
When a rounded estimate is reported, its rounding error should be added to the truncation bound.
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Use |S−r|≤|S−S_N|+|S_N−r|.
Find the exact rounding difference $|5/6-0.833|$.
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The difference is 1/3000, separate from the truncation error.
Rounding an interval lower endpoint upward and its upper endpoint downward always preserves a guaranteed enclosure.
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These inward changes may exclude the true sum. Round the lower endpoint down and the upper endpoint up.
Eventual decrease can still give a valid bound
- For magnitudes $b_n=n/(n^2+9)$, $f'(x)=(9-x^2)/(x^2+9)^2\le0$ from $x=3$ and $b_n\to0$. After the first two terms, the omitted tail starts at $3$ and satisfies the conditions.
- Thus its alternating series has an error bound $b_3=1/6$ after two terms, despite the early magnitude increases. Check where the omitted tail begins before using the theorem.
If two terms have been included from index 1, what is the first omitted index?
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For b_n=n/(n²+9), this is the start of a valid non-increasing omitted tail.
Practise certificates, indexing and rounding
- For magnitudes $2/(n+1)^2$ starting at $n=1$, find the smallest $N$ satisfying the next-term conditions for error at most $0.005$ and strictly below $0.005$.
- For $\sum_{n=4}^{\infty}(-1)^{n-4}/n$, identify the omitted index after five terms and give an enclosure. Then combine a truncation bound $0.004$ with rounding to three decimal places.
For alternating magnitudes $b_n=2/(n+1)^2$, starting at n=1, find the smallest N satisfying the next-term conditions for error at most 0.005 and strictly below 0.005.
Compare the term count, omitted index and stated guarantee with the model.
For $\sum_{n=4}^{\infty}(-1)^{n-4}/n$, give the omitted index, error bound and enclosure after five terms.
Compare the term count, omitted index and stated guarantee with the model.
An exact partial sum has truncation error at most 0.004. It is rounded to the nearest three decimal places. Give a combined bound and say whether it certifies error at most 0.005.
Compare the term count, omitted index and stated guarantee with the model.
State what is guaranteed
- Verify the omitted-tail conditions, identify the next index and bound the error by its magnitude. State whether the target uses $\le$ or $<$, and say how many terms are included.
- Keep exact partial sums, rounded estimates and actual errors separate. Add a rounding allowance when needed; a small-looking finite plot is not an error certificate.
For a valid alternating omitted tail, the exact sum is between two consecutive exact partial sums and the error is at most the first omitted magnitude. Count indices carefully, use a condition that certifies the requested inequality, and include any rounding error in the reported estimate.