Check that the alternating error bound applies

Partial sums on both sides of the sum

The series is \(\tfrac13-\tfrac19+\tfrac1{27}-\cdots\). Term \(n\) has size \(1/3^n\), and the signs alternate.

Each teal dot is a partial sum \(S_n\), the total of the first \(n\) terms. A green segment is a step up, and a red segment is a step down. The orange bar under a dot shows the size of the step that reaches it.

At 2 terms, the first dot is \(S_1=0.333\), and the second term steps down by \(\tfrac19\) to \(S_2=0.222\). The orange ring marks the next partial sum, \(S_3=0.259\), one step of \(\tfrac1{27}\) up.

Each step is shorter than the last and reverses direction, so the dots lie on both sides of the dashed line at the sum, 0.25.

The sum lies in the faint strip from \(S_2\) to the ring. This means \(S_2\) is closer to the sum than the next step is long, 0.028 against \(\tfrac1{27}=0.037\).

Any alternating series

For any alternating series, write the size of term \(n\) as \(b_n\), so the steps have the lengths \(b_1, b_2, b_3, \dots\).

Each step goes the opposite way to the step before it and is shorter, so the sum lies between any two neighboring dots.

This means \(S_n\) is closer to the sum than the length \(b_{n+1}\) of the next step, which the orange bracket shows.

If a step were longer than the step before it, the sum could lie outside the faint strip.

Notation

1. The error of a partial sum is the amount that it misses: the sum \(S\) minus the partial sum \(S_n\). The red bracket shows it.

   \[\cr{S-S_n}\]

2. The alternating error bound says that the size of the error is at most the size of the next term, the first omitted term. The orange bracket shows its step.

   \[|S-S_n|\le\cy{b_{n+1}}\]

3. The bound needs three conditions. The signs alternate, each size is no larger than the size before it, and the sizes approach 0.

   \[b_{k+1}\le b_k,\qquad \lim_{k\to\infty}b_k=0\]

4. The sizes need to decrease only from the first omitted term on, because the earlier terms are already in \(S_n\).

   If the sizes decrease from term \(N\) on, the bound holds for \(S_n\) when \(n+1\ge N\).

When it fails

Make the sizes uneven: \(1/n^3\) for odd \(n\) and \(1/n^2\) for even \(n\). The sizes are \(1, \tfrac14, \tfrac1{27}, \tfrac1{16}, \dots\). The signs alternate, and the sizes approach 0.

But \(b_3=\tfrac1{27}\) is smaller than \(b_4=\tfrac1{16}\). The sum is 0.641, and \(S_3=0.787\) is 0.146 away from it. The next step is 0.0625, so the bound fails.

Example 1: the conditions hold

For \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2+2}\), decide whether the alternating error bound applies to \(S_4\). If it does, state the bound.

1. The factor \((-1)^{n+1}\) makes the signs alternate. The sizes are \(b_n=\dfrac1{n^2+2}\).

2. The denominator increases with \(n\), so each size is smaller than the one before. The arrows join each bar to the next, shorter bar.

   \[b_{n+1}

3. The denominator grows without limit, so the sizes approach 0.

   \[\lim_{n\to\infty}\cy{\frac1{n^2+2}}=0\]

4. All three conditions hold, so the bound applies. The first omitted term after \(S_4\) has \(n=5\).

   \[|S-S_4|\le b_5=\frac1{27}\]

Example 2: the conditions hold only from some term on

For \(\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{n}{n^2+16}\), find the smallest \(n\) for which the alternating error bound applies to \(S_n\).

1. The first sizes are \(b_1=0.059\), \(b_2=0.100\), \(b_3=0.120\), \(b_4=0.125\) and \(b_5=0.122\), so the sizes grow at first and then fall.

2. Let \(f(x)=\dfrac{x}{x^2+16}\), so \(b_n=f(n)\). Its derivative is negative when \(x>4\), so the sizes decrease from \(n=4\) on.

   \[f'(x)=\frac{16-x^2}{(x^2+16)^2}\]

3. The sizes approach 0, because the denominator grows faster than the numerator.

   \[\lim_{n\to\infty}\cy{\frac{n}{n^2+16}}=0\]

4. A tempting wrong move: The signs alternate and the sizes approach 0, so the bound holds for \(S_2\).

   \[|S-S_2|\le b_3\]

   The sizes increase from \(b_3=0.120\) to \(b_4=0.125\), so the step from \(S_3\) to \(S_4\) is longer than the step before it. The red ring shows \(S_4\) below \(S_2\), outside the faint strip, so nothing keeps the sum inside it.

5. The sizes must decrease from the first omitted term on. For \(S_n\) that term has the index \(n+1\), so \(n+1\ge4\), and the smallest \(n\) is 3.

   \[n+1\ge4\;\Rightarrow\;n\ge3\]

6. For \(S_3\), the bound is \(b_4\). The strip between \(S_3\) and \(S_4\) holds the dashed line.

   \[|S-S_3|\le b_4=\frac4{32}=\frac18\]

Practice

For \(\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{n}{2^n}\), show that the alternating error bound applies to \(S_5\), and state the bound.

Each orange bar must be no taller than the one before it. Is \(b_{n+1}/b_n\) at most 1 for every \(n\)?

Questions

Each graph has teal dots for the partial sums of an alternating series. The dashed line marks the sum, and the faint strip joins \(S_3\) to \(S_4\). Green steps go up and red steps go down.

1. In the upper graph, each step is shorter than the one before. What follows?

   • The sum lies between each dot and the next dot.
   • The sum lies above every dot.
   • The sum is equal to the last dot shown.
   Answer

   The sum lies between each dot and the next dot.

2. In the lower graph, the signs alternate and the sizes approach 0. Which condition of the bound fails?

   • The sizes do not always decrease.
   • The signs do not alternate.
   • The sizes do not approach 0.
   Answer

   The sizes do not always decrease.

3. The sizes of a series are \(b_n=\dfrac{n}{n^2+9}\). From which \(n\) on do the sizes decrease?

   • \(n=3\)
   • \(n=1\)
   • \(n=4\)
   • \(n=9\)
   Answer

   \(n=3\)

Summary

You confirm that the signs alternate, that the sizes decrease from the first omitted term on, and that the sizes approach 0.

These conditions make each step shorter than the last and opposite to it, so the error is at most the size of the next term.

You confirm them before you quote an error bound for an alternating partial sum.

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