Check eventual decrease

Jumps that grow, then shrink

The teal dots are the partial sums of an alternating series with sizes \(a_n=\frac{8n}{n^2+9}\). The orange bracket marks the newest jump, and its length is the size of the term that the series adds.

Drag the slider from 1 to 3 terms. The sizes grow from 0.8 to about 1.23 and then to \(\frac43\), so each jump is longer than the one before and the zigzag widens.

Continue to 8 terms. The sizes fall, about 1.28, 1.18 and 1.07, so each jump is shorter.

From the fourth dot on, each new dot lies inside the faint strip between the two dots before it.

The sizes fall only from the third term on, and the test needs no more than that.

Falling from some term on

Drag the marker to a term \(N\). To the right of the marker, each orange bar is no longer than the bar before it.

Then each new partial sum lies between the two sums before it, because the new jump is not longer than the last one.

To the left of the marker, the bars can rise and fall in any way. These first terms add the same fixed amount to every later partial sum, so they do not decide whether the series converges.

The condition is that, from some term \(N\) on, the sizes do not increase.

Notation

1. The second condition of the alternating series test says that, from some term \(N\) on, each size is not larger than the size before it.

   \[\cy{a_{n+1}}\le\cy{a_n}\quad\text{for all }n\ge N\]

2. Method 1: write \(a_{n+1}\) by replacing \(n\) with \(n+1\), and compare it with \(a_n\). A larger denominator or a smaller numerator gives a smaller size.

3. Method 2: divide \(a_{n+1}\) by \(a_n\). The sizes are positive, so a quotient of at most 1 means that \(a_{n+1}\le a_n\).

   \[\frac{a_{n+1}}{a_n}\le1\]

4. Method 3: find a function \(f\) with \(f(n)=a_n\). If \(f'(x)\le0\) for all \(x\ge N\), then \(f\) does not rise there, so neither do the sizes. The curve is green where it rises and red where it falls.

   \[f'(x)\le0\ \text{for }x\ge N\]

When it fails

The sizes can approach 0 and still rise again and again. For \(a_n=\frac{2+(-1)^n}{n}\) the sizes are \(1,\ \frac32,\ \frac13,\ \frac34,\ \frac15,\ \frac12\). Each even size is larger than the odd size before it, and the red rings mark these bars.

No term N exists from which the sizes fall, so the alternating series test gives no conclusion here.

Example 1: Comparing denominators

Show that \(a_{n+1}\le a_n\) for all \(n\ge1\), where \(a_n=\frac{5}{n^2+n}\).

1. Write the next size. Replace \(n\) with \(n+1\) in the formula.

   \[\begin{aligned}a_{n+1}&=\frac{5}{(n+1)^2+(n+1)}\\&=\frac{5}{n^2+3n+2}\end{aligned}\]

2. The difference of the two denominators is positive for \(n\ge1\), so the denominator of \(a_{n+1}\) is larger. At \(n=3\), for example, they are 12 and 20.

   \[\begin{aligned}&(n^2+3n+2)-(n^2+n)\\&=2n+2>0\end{aligned}\]

3. Both numerators are 5, and a larger denominator gives a smaller fraction. So \(a_{n+1}\lt a_n\) for every \(n\ge1\), and \(N=1\).

Example 2: Using a derivative

For the series \(\sum_{n=2}^{\infty}(-1)^n\frac{\ln n}{n}\), find the term from which the sizes \(a_n=\frac{\ln n}{n}\) fall.

1. The size at \(n=3\) is larger than the size at \(n=2\).

   \[a_2\approx0.347,\quad a_3\approx0.366\]

2. A tempting wrong move: The size grew from \(n=2\) to \(n=3\), so the sizes do not fall, and the condition fails.

   The condition asks for a fall from some term on, not from the first term. In the graph the bars rise up to \(n=3\), and every later bar is shorter.

3. Use \(f(x)=\frac{\ln x}{x}\), which has \(f(n)=a_n\). Its derivative is negative when \(\ln x>1\), that is, for \(x>e\approx2.72\).

   \[f'(x)=\frac{1-\ln x}{x^2}\]

4. Because \(3>e\), the derivative is negative for \(x\ge3\). So \(a_{n+1}\le a_n\) for all \(n\ge3\), and \(N=3\).

Practice

Find the smallest \(N\) such that \(a_{n+1}\le a_n\) for all \(n\ge N\), where \(a_n=\frac{(5/2)^n}{n!}\).

Divide \(a_{n+1}\) by \(a_n\). The sizes do not grow once the quotient is at most 1.

Questions

The teal dots are the partial sums of \(\sum(-1)^na_n\) for \(n=1\) to 8, and the orange bars are the sizes \(a_n=n(0.7)^n\).

1. What is the smallest \(N\) such that \(a_{n+1}\le a_n\) for every \(n\ge N\)?

   • 3
   • 1
   • 2
   • 4
   Answer

   3

2. What does the graph show about the partial sums from the fourth dot on?

   • Each new dot lies between the two dots before it.
   • Each jump is longer than the jump before it.
   • The dots move in one direction.
   Answer

   Each new dot lies between the two dots before it.

3. For \(\sum(-1)^n\frac{n^2}{3^n}\), what is the smallest \(N\) such that \(a_{n+1}\le a_n\) for all \(n\ge N\), where \(a_n=\frac{n^2}{3^n}\)?

   • 2
   • 1
   • 3
   • 4
   Answer

   2

Summary

You show that, from some term on, each size is not larger than the one before, by a comparison, a quotient or a derivative.

This works because a shorter jump brings each new partial sum back between the two before it, so the zigzag narrows.

You use it as the second condition of the alternating series test, and the first terms do not matter.

This is one of four skills in topic 10.7 of Which & Why: AP Calculus.