Recognize the inconclusive zero case

Terms that shrink, a total that grows

The harmonic series is \(1+\frac12+\frac13+\frac14+\cdots\). In the upper graph, each bar is one term \(\frac1n\), so the bars shrink toward 0. In the lower graph, each dot is a partial sum, the total of the terms up to that \(n\).

Press the button to add the next block of bars, which ends at \(n=4\), 8 or 16.

Each new block adds at least \(\frac12\), as the orange bracket shows. Bars 3 and 4 are each at least \(\frac14\) tall, so they add at least \(2\cdot\frac14=\frac12\). Bars 5 to 8 are each at least \(\frac18\) tall, so the four add at least \(4\cdot\frac18=\frac12\).

The blocks never end, so the total becomes larger than any number, even though the terms shrink to 0.

The bars shrink, and the number of bars grows

Drag the slider to show more or fewer terms. Each block holds twice as many bars as the block before it, and its bars are about half as tall.

The two changes balance, so every block adds at least \(\frac12\), as the orange brackets show. The total grows without end, although the bars shrink to 0.

Notation

1. If a series converges, its terms approach 0. The reverse is not true.

   \[\sum a_n\ \text{converges}\ \Rightarrow\ \lim_{n\to\infty}a_n=0\]

2. The harmonic series has terms that approach 0, but it diverges, so a zero limit of the terms does not prove convergence.

   \[\begin{aligned}\lim_{n\to\infty}\tfrac1n&=0\\ \cr{1+\tfrac12+\tfrac13+\cdots}&\ \cr{\text{diverges}}\end{aligned}\]

3. For \(k\ge1\), block \(k\) holds the \(2^k\) terms from \(\frac1{2^k+1}\) to \(\frac1{2^{k+1}}\). Each is at least \(\frac1{2^{k+1}}\), so the block adds at least \(\frac12\). The first two terms add \(\frac32\), so the partial sums have no upper bound.

   \[S_{2^k}\ge1+\tfrac{k}{2}\]

4. If the limit of the terms is 0, the nth-term test is inconclusive, which means it does not decide. You need another test.

When it fails

Replace the terms by \((\frac12)^n\). They also approach 0, but the bars for \(n=3\) and \(n=4\) now add only \(\frac3{16}\), and each later block adds less.

The dots stay below the faint dashed line at height 1, so this series converges. Convergence depends on how fast the terms shrink, not only on whether they shrink.

Example 1: Terms that shrink, and the series converges

The terms of \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}\) approach 0. Find the partial sum \(S_N\) and decide whether the series converges.

1. Split each term into two fractions. Over the common denominator \(n(n+1)\), the right side equals the left side.

   \[\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}\]

2. Add the first \(N\) terms. The second fraction of each term cancels the first fraction of the next term, so only the first and the last pieces remain.

   \[\begin{aligned}S_N&=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots\\ &\quad+\left(\tfrac1N-\tfrac1{N+1}\right)\\ &=1-\tfrac1{N+1}\end{aligned}\]

3. The teal dots approach the faint dashed line at height 1, so this series converges.

   \[\lim_{N\to\infty}S_N=1\]

Example 2: Terms that shrink, and the series diverges

Decide whether \(\displaystyle\sum_{n=1}^{\infty}\left(\sqrt{n+1}-\sqrt{n}\right)\) converges.

1. Multiply and divide the term by \(\sqrt{n+1}+\sqrt{n}\). The top becomes \((n+1)-n=1\) and the bottom grows without bound, so the terms approach 0.

   \[\begin{aligned}\sqrt{n+1}-\sqrt n&=\frac{1}{\sqrt{n+1}+\sqrt n}\\ &\to0\end{aligned}\]

2. A tempting wrong move: The terms approach 0, as in Example 1, so the series converges.

   \[\lim_{n\to\infty}a_n=0\ \Rightarrow\ \text{converges}\]

   The red circle marks the sum of the first 30 terms, about 4.57, and the teal bars continue to rise. After 100 terms the sum is about 9.05.

3. Add the first \(N\) terms. Each \(\sqrt{n+1}\) cancels the \(-\sqrt{n+1}\) in the next term. The sum that remains grows without end, so the series diverges.

   \[\begin{aligned}S_N&=\left(\sqrt2-1\right)+\left(\sqrt3-\sqrt2\right)\\ &\quad+\cdots+\left(\sqrt{N+1}-\sqrt N\right)\\ &=\cb{\sqrt{N+1}-1}\end{aligned}\]

Practice

Show that the terms of \(\displaystyle\sum_{n=1}^{\infty}\ln\!\left(1+\dfrac1n\right)\) approach 0. Then decide whether the series converges.

Write \(\ln(1+\frac1n)\) as \(\ln(n+1)-\ln n\). Then add the first \(N\) terms and compare the result with the lower graph.

Questions

Questions 1 and 2 use the figure. The upper graph has bars for the terms \(\frac1n\) with \(n=1\) to 16, and the bars for \(n=5\) to 8 are teal. The lower graph has the partial sums of two series whose terms approach 0.

1. The shortest bar in the block \(n=5\) to 8 has height \(\frac18\). What does this say about the total of the block?

   • The block adds at least \(\frac12\).
   • The block adds at least \(\frac18\).
   • The block adds at most \(\frac12\).
   Answer

   The block adds at least \(\frac12\).

2. Both series have terms that approach 0. What do the two sets of dots show?

   • Terms that approach 0 can go with partial sums that approach a height, and also with partial sums that rise without end.
   • Terms that approach 0 always give partial sums that approach a height.
   • Terms that approach 0 always give partial sums that rise without end.
   Answer

   Terms that approach 0 can go with partial sums that approach a height, and also with partial sums that rise without end.

3. The terms of \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+2)}\) approach 0. What is the sum of the series?

   • \(\tfrac34\)
   • \(\tfrac32\)
   • 1
   • The series diverges.
   Answer

   \(\tfrac34\)

Summary

The bars of the harmonic series shrink to 0, but each block of them adds at least \(\frac12\), so the total grows without end.

A zero limit of the terms is necessary for convergence, but it does not prove convergence.

When the terms of a series approach 0, you write that the nth-term test is inconclusive, and you use another test.

This is one of four skills in topic 10.3 of Which & Why: AP Calculus.