Recognize harmonic divergence
A total that grows very slowly
The harmonic series is \(1+\frac12+\frac13+\frac14+\cdots\). Its terms are \(\frac1n\), so it is the p-series with \(p=1\).
The teal dots show the running total \(S_N\), the sum of the first \(N\) terms, for \(N=10\), 100, 1000 and 10 000. Each step to the right on the horizontal axis means ten times as many terms.
Move the slider, and the orange dot moves to the chosen N. The totals are about 2.93, 5.19, 7.49 and 9.79, while the last term shrinks from 0.1 to 0.0001. Each time N becomes ten times larger, the total grows by about 2.3.
The growth is slow, but it does not stop. The next scene shows why the total passes every bound.
Blocks that each add at least one half
The bars show the terms \(\frac1n\), now without numbers. The first two bars stand alone. The later bars form blocks of 2, 4 and 8 bars, in alternating colors.
Press the first button. Each bar drops to the height of the last bar in its block, so each block total becomes smaller. A faint outline keeps the old height.
A block has twice as many bars as the one before, and its last bar is half as tall. The two changes cancel, so the lowered total of every block is the same number, \(\frac12\).
Notation
Let \(S_N\) be the total of the first \(N\) terms of the harmonic series.
\[S_N=1+\frac12+\frac13+\cdots+\frac1N\]The first two terms add to \(\frac32\). The next block has 2 terms, and each is at least \(\frac14\), the last term of the block.
\[\begin{aligned}S_2&=\frac32\\ \frac13+\frac14&>\frac14+\frac14=\cy{\frac12}\end{aligned}\]The next block has 4 terms, and each is at least \(\frac18\). The block after that has 8 terms, and each is at least \(\frac1{16}\).
\[\begin{aligned}\frac15+\cdots+\frac18&>\frac48=\cy{\frac12}\\ \frac19+\cdots+\frac1{16}&>\frac8{16}=\cy{\frac12}\end{aligned}\]For \(k\ge2\), the first \(2^{k}\) terms are the first two terms and \(k-1\) blocks. Each block adds more than \(\frac12\), which gives the lower bound below, called the block bound.
\[S_{2^{k}}>\frac32+\frac{k-1}{2}=1+\frac{k}{2}\]If you choose \(k\) large enough, \(1+\frac{k}{2}\) passes any bound \(M\), and so does \(S_{2^{k}}\). The totals have no bound, so the harmonic series diverges.
\[\sum_{n=1}^{\infty}\frac1n=\infty\]
When it fails
The terms \(\frac1n\) approach 0, so you might expect the total to end its growth. It does not, because each block adds more than \(\frac12\), however small the bars become.
Switch the bars to \(\frac1{n^{2}}\). The lowered blocks now add only \(\frac18\), \(\frac1{16}\) and \(\frac1{32}\), and each is half of the one before, so together they stay below \(\frac14\). They no longer prove divergence, and this p-series converges because \(p=2>1\).
Example 1: The first sixteen terms
Show that \(1+\frac12+\frac13+\cdots+\frac1{16}\) is more than 3.
The first two terms add to \(\frac32\).
\[1+\frac12=\frac32\]Every term of a block is at least as large as its last term, so each of the three blocks adds more than \(\frac12\).
\[\begin{aligned}\frac13+\frac14&>\frac24\\ \frac15+\cdots+\frac18&>\frac48\\ \frac19+\cdots+\frac1{16}&>\frac8{16}\end{aligned}\]Add the three bounds to the first two terms. As a check, the actual sum is about 3.38.
\[S_{16}>\frac32+\cy{\frac12}+\cy{\frac12}+\cy{\frac12}=3\]
Example 2: How many terms pass 6?
Use the blocks to find a number of terms that makes the harmonic total more than 6.
The orange dots show the block bound \(1+\frac{k}{2}\) at \(N=2^{k}=4\), 8, 16 and so on, just below the teal totals.
\[S_{2^{k}}>1+\frac{k}{2}\]A tempting wrong move: The first 10 terms add to about 2.93. Doubling the number of terms doubles the total, so 20 terms give about 5.9.
\[S_{20}\approx 2\cdot 2.93=5.9\]The teal dot shows the real total for 20 terms, about 3.60, far below the red dot. Doubling the terms from 10 to 20 adds only about 0.67.
Each doubling of the number of terms adds more than \(\frac12\). The bound \(1+\frac{k}{2}\) reaches the faint dashed line at 6 when \(k=10\).
\[1+\frac{10}{2}=6\]The first \(2^{10}=1024\) terms add to more than 6. As a check, the actual total is about 7.51.
\[S_{1024}>6\]
Practice
Use the blocks to show that the first 256 terms of the harmonic series add to more than 5.
Write 256 as \(2^{k}\). Each bar shows the total that the block bound \(1+\frac{k}{2}\) guarantees.
Solution
Write the number of terms as a power of 2.
\[256=2^{8}\]Use the block bound with \(k=8\).
\[S_{256}>1+\frac{8}{2}=\cb{5}\]The first 256 terms add to more than 5.
Questions
The bars show the terms \(\frac1n\) for \(n\) from 1 to 16. The orange bars are one block of terms, and the dashed line marks the height of the last bar of that block.
Every orange bar is at least as tall as the dashed line. What does this show about the total of the four orange bars?
- It is more than \(\frac12\).
- It is more than \(\frac18\).
- It is more than 1.
Answer
It is more than \(\frac12\).
The orange bars are now the eight bars from \(n=9\) to \(n=16\). How does their guaranteed total compare with that of the bars from \(n=5\) to \(n=8\)?
- It is the same, \(\frac12\).
- It is half as large, because the bars are shorter.
- It is twice as large, because there are more bars.
Answer
It is the same, \(\frac12\).
The block bound says \(S_{2^{k}}>1+\frac{k}{2}\). Which of these numbers of terms is the smallest one that guarantees a total above 8?
- 16 384
- 8192
- 65 536
- 256
Answer
16 384
Summary
You group the terms of \(\frac1n\) into blocks of 2, 4, 8 and more terms, and each block adds more than \(\frac12\).
The number of blocks has no limit, so the totals pass every bound and the harmonic series diverges.
A p-series with p = 1 is the harmonic series, so it diverges even though its terms approach 0.
This is one of four skills in topic 10.5 of Which & Why: AP Calculus.