Check that magnitudes approach zero
How far each partial sum jumps
The upper graph shows the partial sums of an alternating series as teal dots joined by a zigzag line. A partial sum is the total of the first \(n\) terms.
Series A has sizes \(a_n=\frac{3n}{2n+1}\), where a size is a term without its sign. Its first partial sums are \(1\), \(-\frac15\), \(\frac{38}{35}\) and \(-\frac{26}{105}\), so the dots jump up and down at every step.
The orange bracket shows the last jump, and its length is the size of the term added at that step.
In series A the sizes approach \(\frac32\), so the later jumps are all about \(\frac32\) long and the dots cannot approach one value.
Press the button to use series B, with sizes \(\frac{3}{2n+1}\). Its first jumps are \(1\), \(\frac35\), \(\frac37\) and \(\frac13\), so the jumps become shorter and the dots move toward one value.
Where the sizes go
Drag the first slider to change the number \(L\) that the sizes \(a_n\) approach. The dashed orange line in the lower graph marks \(L\).
The jump from one partial sum to the next is as long as the size of the new term.
If \(L\) is 0, the jumps become as short as you like, so the dots can come together at one value.
If \(L\) is more than 0, the jumps stay about \(L\) long, so the dots do not come together and the series diverges.
The limit of the sizes is therefore the first fact to test.
Notation
The alternating series test has two conditions, and the first says that the sizes approach 0.
\[\lim_{n\to\infty}\cy{a_n}=0\]If \(a_n\) is a fraction, divide the top and the bottom by the highest power of \(n\) in the bottom. Then every term of the form \(c/n\) approaches 0.
\[\frac{3n}{2n+1}=\frac{3}{2+\cq{1/n}}\to\frac32\]The limit is 3/2, not 0. This condition fails for series A.
A series diverges if its terms do not approach 0, so an alternating series diverges if the limit of its sizes \(a_n\) is not 0.
If the limit is 0, this condition holds, but the series can still diverge if the second condition fails.
When it fails
If the sizes approach a number \(L\) other than 0, the terms are close to \(L\) and \(-L\) in turn. They do not approach 0, so the series diverges.
The sizes may also grow without bound. For \(1-2+3-4+\cdots\) the partial sums are \(1\), \(-1\), \(2\), \(-2\) and so on, and the jumps become longer.
Example 1: Sizes that shrink to zero
Does the size \(a_n=\frac{4n+1}{n^2+3}\) of the series \(\sum(-1)^na_n\) approach 0?
The highest power of \(n\) in the bottom is \(n^2\). Divide the top and the bottom by \(n^2\).
\[\frac{4n+1}{n^2+3}=\frac{4/n+1/n^2}{1+3/n^2}\]When \(n\) is large, \(4/n\), \(1/n^2\) and \(3/n^2\) are all close to 0, so the top approaches 0 and the bottom approaches 1.
\[\lim_{n\to\infty}a_n=\frac{0}{1}=0\]The limit is 0, so the orange bars and the jumps shrink toward 0. This condition holds.
Example 2: Sizes that do not shrink
Does the series \(\sum(-1)^n\frac{5n^2+1}{2n^2+n}\) converge?
Divide the top and the bottom by \(n^2\), the highest power in the bottom.
\[\frac{5n^2+1}{2n^2+n}=\frac{5+1/n^2}{2+1/n}\]The top approaches 5 and the bottom approaches 2, so the sizes approach \(\frac52\). The orange bars level off at the dashed line.
\[\lim_{n\to\infty}a_n=\frac52\]A tempting wrong move: The signs alternate, so the positive terms and the negative terms cancel, and the series converges.
The orange bracket is still about \(\frac52\) long, so the partial sums jump down and up at every step and do not come together.
The sizes approach \(\frac52\), not 0, so the terms do not approach 0 and the series diverges.
Practice
Does the size \(a_n=\frac{n^2+2}{3n+1}\) approach 0? What does your answer say about \(\sum(-1)^na_n\)?
Move the slider to a large value of \(n\). Do the orange bars become shorter or longer?
Solution
The highest power of \(n\) in the bottom is \(n\). Divide the top and the bottom by \(n\).
\[\frac{n^2+2}{3n+1}=\frac{n+2/n}{3+1/n}\]The top grows without bound and the bottom approaches 3, so the sizes grow without bound, as the taller orange bars show.
\[\lim_{n\to\infty}a_n=\infty\]The limit is not 0, so the terms do not approach 0. The series diverges.
Questions
The graph shows the partial sums of an alternating series with sizes \(a_n=\frac{4n}{2n+1}\), for \(n=1\) to 6. The orange bars show the sizes.
What number do the sizes (the orange bars) approach?
- 2
- 0
- \(\tfrac43\)
- They grow without bound.
Answer
2
What does the zigzag show about the partial sums?
- They keep jumping between about 1.5 and −0.3, so they do not approach one number.
- They approach the middle value, about 0.6.
- They approach 0, because the terms alternate.
Answer
They keep jumping between about 1.5 and −0.3, so they do not approach one number.
For which sizes \(a_n\) does \(\sum(-1)^na_n\) diverge because \(a_n\) does not approach 0?
- \(a_n=\frac{6n}{3n+7}\)
- \(a_n=\frac{n+3}{n^2+1}\)
- \(a_n=\frac{5}{\sqrt{n+2}}\)
- \(a_n=\frac{3n}{n^2+2n}\)
Answer
\(a_n=\frac{6n}{3n+7}\)
Summary
You find the limit of the sizes \(a_n\). For a fraction, you divide the top and the bottom by the highest power of \(n\) in the bottom.
This works because the dots can approach one value only if the jumps, which are the sizes, shrink to 0.
You test this first, because a limit other than 0 shows that the series diverges.
This is one of four skills in topic 10.7 of Which & Why: AP Calculus.