Is S_3 too large or too small as an approximation of the sum?

It cannot be decided from the picture

Let S_3 approximate \sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2+1}. Which statement is correct?

S_3 is too large, and the error is at most \tfrac1{17}.

S_3 is too small, and the error is at most \tfrac1{17}.

S_3 is too large, and the error is at most \tfrac1{10}.

S_3 is too small, and the error is at most \tfrac1{10}.

Topic 10.10 · BC

Bound the size and direction of the error

Builds on identifying the first omitted term. 7 minutes.

Above or below the sum

The series is \(1-\tfrac14+\tfrac19-\tfrac1{16}+\cdots\). Term \(n\) has size \(1/n^2\), and the sum, the dashed line, is about 0.8225.

The teal dots are the partial sums, joined by green steps up and red steps down. The first dot, \(S_1=1\), is above the dashed line, \(S_2=0.75\) is below it, and \(S_3=0.861\) is above it again.

Here, the partial sums with an odd number of terms are above the sum, and the others are below it.

At 5 terms, the orange dot is \(S_5=0.839\). The next term is \(-\tfrac1{36}\), so the next step goes down 0.028 to \(S_6=0.811\).

The orange bracket marks this step, and the sum lies inside it. So \(S_5\) is above the sum, and less than 0.028 from it.

The next step shows both

Take an alternating series whose sizes decrease to 0. Its sum, the dashed line, lies inside the next step from \(S_n\) to \(S_{n+1}\), which the orange bracket marks.

This step has the length \(\cy{b_{n+1}}\), the size of the first omitted term. So the size of the error, the distance from \(S_n\) to the sum, is at most \(b_{n+1}\).

The next step points toward the sum, and it has the sign of the first omitted term. If that term is positive, the sum is above \(S_n\). If it is negative, the sum is below \(S_n\). This sign is the direction of the error.

Notation

The error of \(S_n\) is the sum minus the partial sum, and the red bracket shows it.
\[\cr{E_n}=S-S_n\]
The size of the error is at most \(\cy{b_{n+1}}\), the size of the first omitted term, which is the length of the orange bracket.
\[|E_n|\le\cy{b_{n+1}}\]
The error has the same sign as the first omitted term \(a_{n+1}\), so this sign gives the direction of the error.
\[E_n\text{ and }a_{n+1}\text{ have the same sign}\]
Both facts hold because the sum lies in the faint strip between \(S_n\) and the next partial sum.
\[\begin{aligned}S&\text{ lies between }S_n\text{ and }S_{n+1},\\ S_{n+1}&=S_n+a_{n+1}\end{aligned}\]
If \(E_n\) is positive, \(S_n\) is too small, an underestimate. If \(E_n\) is negative, \(S_n\) is too large, an overestimate.
These statements need three conditions: the signs alternate, the sizes decrease, and the sizes approach 0.

When it fails

Change the sign of every term. The picture turns upside down, so every partial sum with an odd number of terms, starting with the first dot, is below the sum.

The rule from the opening, that odd partial sums are above the sum, fails. The sign of the first omitted term still works: after \(S_1\) the next term is positive, so the sum is above \(S_1\).

Example 1: a partial sum that is too large

Use \(S_3\) to approximate \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{(2n-1)^2}\). Bound the error, and say whether \(S_3\) is too large or too small.

Add the first three terms.
\[\begin{aligned}\cy{S_3}&=1-\frac19+\frac1{25}\\ &=\frac{209}{225}\approx0.9289\end{aligned}\]
The first omitted term has \(n=4\). It is negative.
\[\cr{a_4}=\frac{(-1)^5}{7^2}=-\frac1{49}\]
The signs alternate, and the sizes \(1/(2n-1)^2\) decrease to 0, so the bound applies.
The error is at most the size of the first omitted term.
\[|S-S_3|\le\frac1{49}\approx0.0204\]
The first omitted term is negative, so the next step goes down. The sum is below \(S_3\), so \(S_3\) is too large.
\[\begin{aligned}S_4&\approx0.9085\le S\\ &\le S_3\approx0.9289\end{aligned}\]

Example 2: a partial sum that is too small

Use \(S_4\), the sum of the first four terms, to approximate \(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2^n(n+1)}\). Bound the error, and say whether \(S_4\) is too large or too small.

Add the first four terms, which have the indexes 0 to 3.
\[\begin{aligned}\cy{S_4}&=1-\frac14+\frac1{12}-\frac1{32}\\ &=\frac{77}{96}\approx0.8021\end{aligned}\]
The first omitted term has \(n=4\). It is positive.
\[\cg{a_4}=\frac{(-1)^4}{2^4\cdot5}=\frac1{80}\]
A tempting wrong move: The last term added is negative, so \(S_4\) is above the sum. It is too large.
\[a_3=-\frac1{32}<0\;\Rightarrow\;S
The dashed red bracket marks the last step, which is already part of \(S_4\). The next step points toward the sum, and it is green because it goes up, so the sum is above \(S_4\).
The direction comes from the first omitted term, not the last term added. That term is positive, so the sum is above \(S_4\), and \(S_4\) is too small.
The error is at most \(1/80\).
\[S_4\;\le\;S\;\le\;S_4+\frac1{80}\approx0.8146\]

Practice

Use \(S_5\) to approximate \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n}\). Bound the error, and say whether \(S_5\) is too large or too small.

Find the first omitted term and its sign. The next step points toward the sum.

Solution

Add the first five terms.
\[\begin{aligned}\cy{S_5}&=\frac12-\frac14+\frac16-\frac18+\frac1{10}\\ &=\frac{47}{120}\end{aligned}\]
The first omitted term has \(n=6\). It is negative.
\[\cr{a_6}=\frac{(-1)^7}{2\cdot6}=-\frac1{12}\]
The error is at most \(1/12\). The omitted term is negative, so the sum is below \(S_5\), and \(S_5\) is too large.
\[\frac{47}{120}-\frac1{12}\;\le\;S\;\le\;\frac{47}{120}\]

Questions

The teal dots are the partial sums of the series \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2^n}\). The dashed line marks its sum, \(\tfrac13\). Orange brackets mark the steps that reach \(S_4\) and \(S_5\).

Is \(S_3\) too large or too small as an approximation of the sum?
- Too large
- Too small
- It cannot be decided from the picture
Answer
Too large
What is the greatest error that the alternating error bound allows for \(S_4\)?
- \(\tfrac1{32}\)
- \(\tfrac1{16}\)
- 0.021
Answer
\(\tfrac1{32}\)
Let \(S_3\) approximate \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2+1}\). Which statement is correct?
- \(S_3\) is too large, and the error is at most \(\tfrac1{17}\).
- \(S_3\) is too small, and the error is at most \(\tfrac1{17}\).
- \(S_3\) is too large, and the error is at most \(\tfrac1{10}\).
- \(S_3\) is too small, and the error is at most \(\tfrac1{10}\).
Answer
\(S_3\) is too large, and the error is at most \(\tfrac1{17}\).

Summary

You bound the error by the size of the first omitted term, and its sign shows which side of the partial sum holds the sum.

This works because the sum lies between consecutive partial sums, and the next step points toward it.

You use it to say how close an alternating partial sum is, and whether it is too large or too small.

This is one of four skills in topic 10.10 of Which & Why: AP Calculus.