Identify the first omitted term
Three terms of the series for sin 1
The series for \(\sin 1\) is \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)!}\). The symbol \(5!\) means \(1\cdot2\cdot3\cdot4\cdot5=120\), so the series begins \(1-\tfrac16+\tfrac1{120}-\tfrac1{5040}+\cdots\).
Each card shows one term, and its index \(n\) is under the card. The first card has \(n=0\), because the sum starts at \(n=0\).
Move the slider. The teal brace covers the terms that the partial sum adds. The orange card after it is the first omitted term, the first term that the sum leaves out.
At 3 terms, the brace covers \(n=0\), \(1\) and \(2\), because the count starts at 0. Their sum is \(1-\tfrac16+\tfrac1{120}=\tfrac{101}{120}\approx0.8417\).
The first omitted term is the next card, with \(n=3\) and the value \(-\tfrac1{5040}\).
Count by position
The numbers under these cards are positions, not indexes.
A partial sum \(S_k\) adds the first \(k\) terms, so it ends at position \(k\), and the first omitted term is in position \(k+1\).
The index \(n\) in the formula can start at 0 or at 1, so count terms by position, not by the value of \(n\).
Press the button. The names on the cards change, but the brace covers the same cards, and the orange card is still the one directly after it.
Notation
The partial sum \(\cb{S_k}\) is the sum of the first \(k\) terms. In this series the first index is 0, so the last index is \(k-1\).
\[\cb{S_k}=\sum_{n=0}^{k-1}\frac{(-1)^n}{(2n+1)!}\]The first omitted term \(\cy{a_k}\) is the next term, so its index is \(n=k\).
\[\cy{a_k}=\frac{(-1)^k}{(2k+1)!}\]If a series starts at \(n=1\), every index is 1 larger. Then the first omitted term of \(S_k\) has the index \(n=k+1\).
When the alternating error bound applies, the error of \(S_k\) is at most the size of the first omitted term.
\[|S-\cb{S_k}|\le|\cy{a_k}|=\frac1{(2k+1)!}\]For \(S_3\), the bound is \(1/7!=1/5040\).
When it fails
A reader may take \(S_3\) to mean the sum of the terms from \(n=0\) through \(n=3\). That sum has four terms, not three.
The brace then covers four cards, and the first omitted card has \(n=4\) and the size \(1/9!=1/362880\). The bound for \(S_3\) is \(1/5040\), which is much larger, so this misreading claims an accuracy that \(S_3\) does not have.
Example 1: a sum of the first four terms
Let \(S_4\) be the sum of the first four terms of \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{4^n(n+1)}\). Find the first omitted term.
List the terms with their indexes, starting at \(n=0\).
\[\begin{aligned}n=0&:\;1,\quad n=1:\;-\tfrac18,\\ n=2&:\;\tfrac1{48},\quad n=3:\;-\tfrac1{256}\end{aligned}\]Four terms use the indexes 0, 1, 2 and 3, so the brace ends at the card with \(n=3\).
\[\cb{S_4}=1-\tfrac18+\tfrac1{48}-\tfrac1{256}\]The first omitted term is the next card, with \(n=4\). It is positive, because \((-1)^4=1\).
\[\cy{a_4}=\frac{(-1)^4}{4^4\cdot5}=\frac1{1280}\]The sizes \(\dfrac1{4^n(n+1)}\) decrease and approach 0, so the error of \(S_4\) is at most \(\dfrac1{1280}\).
Example 2: a sum that ends at a given index
A partial sum of \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(n+2)^2}\) includes every term up to and including the term with \(n=5\). Find the first omitted term.
The brace covers the terms from \(n=0\) through \(n=5\).
A tempting wrong move: The sum has 5 terms, so the first omitted term has \(n=5\).
\[\text{first omitted term: }n=5\]The brace covers six cards: \(n=0, 1, 2, 3, 4\) and \(5\). The card with \(n=5\) is inside the brace, so it is not omitted. The number of terms is the last index plus 1, because the count starts at 0.
The indexes 0 to 5 make six terms, so the next card has \(n=6\).
\[\begin{aligned}n&=0,1,2,3,4,5\\ &\Rightarrow\ \text{first omitted: }n=6\end{aligned}\]Its value is positive, because \((-1)^6=1\).
\[\cy{a_6}=\frac{(-1)^6}{(6+2)^2}=\frac1{64}\]
Practice
Let \(S_7\) be the sum of the first seven terms of \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{5n-3}\). Find the first omitted term and the bound on the error.
This series starts at \(n=1\). Set the slider to 7 terms. Then read the index of the card after the brace.
Solution
The series starts at \(n=1\), so seven terms have the indexes 1 to 7.
\[\cb{S_7}:\;n=1,2,\dots,7\]The first omitted term has the index \(n=8\). Its sign is negative, because \((-1)^9=-1\).
\[\cy{a_8}=\frac{(-1)^9}{5\cdot8-3}=-\frac1{37}\]The sizes \(\dfrac1{5n-3}\) decrease and approach 0, so the error of \(S_7\) is at most the size of this term.
\[|S-S_7|\le\frac1{37}\]
Questions
The cards show the terms of \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(n+3)!}\). The index \(n\) is under each card. In each row, the teal brace covers the terms of a partial sum.
In the upper row, which card is the first omitted term?
- The card with \(n=3\), which equals \(-\tfrac1{720}\)
- The card with \(n=2\), which equals \(\tfrac1{120}\)
- The card with \(n=4\), which equals \(\tfrac1{5040}\)
Answer
The card with \(n=3\), which equals \(-\tfrac1{720}\)
In the lower row, the brace is named "the sum through \(n=3\)". Which card is the first omitted term?
- The card with \(n=4\), which equals \(\tfrac1{5040}\)
- The card with \(n=3\), which equals \(-\tfrac1{720}\)
- The card with \(n=5\), which equals \(-\tfrac1{40320}\)
Answer
The card with \(n=4\), which equals \(\tfrac1{5040}\)
A partial sum of \(\sum_{n=2}^{\infty}\dfrac{(-1)^n}{n^2-1}\) ends with the term \(n=6\). What is the alternating error bound?
- \(\dfrac1{48}\)
- \(\dfrac1{35}\)
- \(\dfrac1{63}\)
Answer
\(\dfrac1{48}\)
Summary
To find the first omitted term, you count the terms of the partial sum by position and take the next term.
This works because the error bound is the size of that next term, so a wrong count gives the bound for a different partial sum.
You need this step each time you quote an alternating error bound, and most of all when the index starts at n = 0.
This is one of four skills in topic 10.10 of Which & Why: AP Calculus.