## Table of Contents

- What is a Harmonic Series?
- Harmonic Number
- Harmonic Series Formula
- Harmonic Series Diverges
- Harmonic Series Examples

In this lesson, learn what a harmonic series is and learn the definition of harmonic number and harmonic formula. Finally, discover the divergence of the harmonic series and see examples.
Updated: 02/23/2022

- What is a Harmonic Series?
- Harmonic Number
- Harmonic Series Formula
- Harmonic Series Diverges
- Harmonic Series Examples

What is a harmonic series? Here is the series:

- {eq}\sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots {/eq}

This is the **harmonic series** math definition. This is the series of rational numbers with numerator one and integer denominator values that progressively increase.

The harmonic series is useful throughout mathematics due to its unique properties:

- The nth partial sum of the series is called a harmonic number. Harmonic numbers are used in many different areas of mathematics.
- The harmonic series diverges and is therefore useful for comparisons and other mathematical processes in calculus.

These properties will be explored later in this lesson.

The harmonic series is most commonly known for its usefulness in music. Since the harmonic series is the only natural scale, it is used as the basis for all tone systems. So, whenever a tone sounds, overtones oscillate with it. However, the sound is simultaneous. This structure is always the same and corresponds to a harmonic series that arises in mathematics. It is difficult to hear the harmonics since they vibrate at the same time as the chord so they sound like a single tone.

Here is an image in Figure *1* showing the harmonic partials (harmonic partial sums) that arise in music as they appear on a music sheet:

The harmonic series is also useful in mathematics for a wide variety of processes. The most important is probably the counterexample to the claim that '"if the limit of the terms in a series is zero, then the series converges."' Clearly, the limit of the terms of this series is zero but, as will be shown later, this series does not converge.

What is a harmonic number? Well, a **harmonic number** is any number that can be represented as a partial sum of the harmonic series. That is, a harmonic number, {eq}H_n {/eq}, is the nth partial sum of the harmonic series:

- {eq}H_n = \sum_{k=1}^n \frac{1}{k} {/eq}.

So, if the harmonic series is stopped at a certain n-value, then the resulting sum is the nth partial sum and a harmonic number.

The final term in the nth partial sum is called the nth term. Since the nth partial sum starts at the first term and ends at the nth term, the resulting sum is called the nth partial sum. Moreover, the sum is called a partial sum because it does not take into account every single term of the series but just a portion of it.

Here are some examples of harmonic numbers:

- The first partial sum with {eq}n=1 {/eq} gives {eq}H_1 = \sum_{k=1}^1 \frac{1}{k} = 1 {/eq}.

- The second partial sum with {eq}n = 2 {/eq} gives: {eq}H_2 = \sum_{k=1}^2 \frac{1}{k} = 1 + \frac{1}{2} = \frac{3}{2} {/eq}.

- The seventh partial sum with {eq}n = 7 {/eq} gives: {eq}H_7 = \sum_{k=1}^7 \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} = \frac{363}{140} {/eq}.

These are all harmonic numbers since they can be represented as partial sums of the harmonic series.

The harmonic series formula is:

- {eq}\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots {/eq}

Now, what is the sum of a harmonic series? It is possible to simply sum the terms of the partial sum. Unfortunately, unlike other series, there is no simple formula for calculating the nth partial sum of a harmonic series. The only way to find the partial sum of a harmonic series is to simply sum the terms of the partial sum:

- {eq}H_{n} = \sum_{k=1}^n \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} {/eq}.

This formula will work for any partial sum of a harmonic series. Ultimately, the sum of harmonic series is given by the sum of the terms of the partial series up to the nth term.

A series, {eq}\sum_{n=1}^{\infty} a_n {/eq}, with nth partial sums denoted by {eq}s_n = \sum_{k=1}^{n} a_k {/eq} can be convergent or divergent:

- If the sequence {eq}\lbrace s_n \rbrace {/eq} is convergent and {eq}lim_{n \rightarrow \infty} s_n = s {/eq} exists and is a real number, then the series is
**convergent**and converges to*s*.

- If the sequence is not convergent, or the limit does not exist or equals infinity, then the series is said to be
**divergent**.

Now, does the harmonic series diverge or does the harmonic series converge? What is the harmonic series convergence? Well, here is the classical proof used by French scholar Nicole Oresme to show that the harmonic series diverges:

**Proof:**

- For this particular series, consider the partial sums {eq}s_2, s_4, s_8, s_16, \dots {/eq} where these partial sums take only an even number of terms.

- Want to show that these partial sums become large.
- Start with the first partial sum: {eq}s_2 = 1 + \frac{1}{2} {/eq}.

- Look for any patterns in the partial sums:
- {eq}s_4 = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) > 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) = 1 + \frac{2}{2} {/eq}.

- {eq}s_8 = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) > 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 + \frac{3}{2} {/eq}

- Using this method, it can be shown that {eq}s_{16} > 1 + \frac{4}{2} {/eq}, {eq}s_{32} > 1 + \frac{5}{2} {/eq}, and {eq}s_{64} > 1 + \frac{6}{2} {/eq}.

- This shows that in general, {eq}s_{2^n} > 1 + \frac{n}{2} {/eq}.

- So, as {eq}n \rightarrow \infty {/eq}, {eq}s_{2^n} \rightarrow \infty {/eq} so the sequence {eq}\lbrace s_{n} \rbrace {/eq} is divergent.

- Therefore, the harmonic series diverges. {eq}\square {/eq}

So, this has shown that the harmonic series is divergent. But, is there another way? Here is another proof using elementary calculus:

**Proof:**

- This proof uses the Integral Test for convergence.
- Consider the harmonic series {eq}\sum_{n=1}^{\infty} \frac{1}{n} {/eq} with {eq}a_n = \frac{1}{n} {/eq} in the series.

- Now, let {eq}a_n = f(n) = \frac{1}{n} {/eq} be a function.

- Note that on the interval {eq}[1, \infty] {/eq}, the function
*f*is continuous, positive, and decreasing. Then, by looking at the improper integral {eq}\int_1^{\infty} f(x) dx {/eq}, the convergence or divergence of the series can be known.

- So, look at {eq}\int_{1}^{\infty} \frac{1}{x} dx {/eq}.

- This integral becomes {eq}\int_{1}^{\infty} \frac{1}{x} dx = \left [ ln \vert x \vert \right ]_1^{\infty} = lim_{b \rightarrow \infty} ln(b) - ln(1) {/eq}

- Now, as {eq}b \rightarrow \infty {/eq}, {eq}ln(b) \rightarrow \infty {/eq}.

- Moreover, {eq}ln(1) = 0 {/eq}.

- So, the value of this integral is {eq}\int_{1}^{\infty} \frac{1}{x} dx = \infty {/eq}.

- Therefore, the harmonic series diverges by the Integral Test. {eq}\square {/eq}

Here is an image in Figure *2* showing this improper integral and the area under the curve:

In the image in Figure *2*, the graph of {eq}\frac{1}{x} {/eq} is shown along with the area under its curve from one to infinity. Since the area under the curve is infinite over this interval, the series associated to this function must also diverge.

Here are some harmonic series examples:

**Example 1:**

What is the fourth partial sum of the harmonic sequence?

**Solution:**

- The nth partial sum of a series is found by taking the sum of the first n-terms of the series. In this case, the third partial sum of the harmonic series is given by: {eq}H_4 = \sum{k=1}^4 \frac{1}{k} {/eq}.

- This partial sum then becomes: {eq}H_4 = \sum{k=1}^4 \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} {/eq}.

- Simplifying this partial sum yields: {eq}H_4 = \frac{25}{12} {/eq}. This is the fourth partial sum of the harmonic series.

**Example 2:**

A *p-series* is a series of the form {eq}\sum_{n=1}^{\infty} \frac{1}{n^p} {/eq} where *p* is any positive real number. At what values of *p* does the *p-series* diverge?

**Solution:**

- Firstly, note that if {eq}p = 1 {/eq}, the series is a harmonic series. So, when {eq}p = 1 {/eq}, the series must diverge since it becomes the harmonic series.

- Firstly, if {eq}p < 0 {/eq}, then {eq}lim_{n \rightarrow \infty} \frac{1}{n^p} = \infty {/eq} so it will diverge.

- If {eq}p = 0 {/eq}, then the series becomes: {eq}\sum_{n=1}^{\infty} \frac{1}{1} = 1 + 1 + 1 + \cdots {/eq}. This will diverge.

- If {eq}p = 1 {/eq}, the series is a harmonic series. So, when {eq}p = 1 {/eq}, the series must diverge since it becomes the harmonic series.

- Finally, consider the final two cases of {eq}p < 1 {/eq} and {eq}p > 1 {/eq}.

- Using the integral test with {eq}\int_{1}^{\infty} \frac{1}{x^p} dx {/eq} these two cases will show that the series converges when {eq}p > 1 {/eq} and diverges when {eq}p < 1 {/eq}.

- So, the p-series converges when {eq}p > 1 {/eq} and diverges when {eq}p \leq 1 {/eq}.

- This example also shows that the harmonic series is a special case of a p-series.

**Example 3:**

Consider the *alternating harmonic series*: {eq}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} {/eq}. This is the harmonic series whose terms are alternating signs. Does this series converge or diverge? If it converges, what does it converge to?

**Solution:**

- Since this is an alternating series, use the alternating series test.
- There are two conditions for the alternating series test. The first is that {eq}a_{n+1} \leq a_n {/eq}, or that the terms in the series are always decreasing. The second is that {eq}lim_{n \rightarrow \infty} a_n = 0 {/eq}.

- So, in the alternating harmonic series, {eq}a_n = \frac{1}{n} {/eq}.

- Look at a_{n+1} \leq a_n in terms of this alternating series: {eq}a_{n+1} = \frac{1}{n+1} {/eq} and {eq}a_{n} = \frac{1}{n} {/eq}. Since {eq}n+1 > n {/eq}, {eq}\frac{1}{n+1} < \frac{1}{n} {/eq} which shows that {eq}a_{n+1} \leq a_{n} {/eq}.

- Next, look at the limit: {eq}lim_{n \rightarrow \infty} \frac{1}{n} = \frac{1}{\infty} = 0 {/eq}.

- So, both conditions are met which means this series is convergent.
- Since this series is convergent, it converges to a particular value. What is that value?
- This step uses the fact that {eq}ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n} {/eq}.

- Letting {eq}x = -1 {/eq} yields: {eq}ln(1 - (-1)) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} {/eq}.

- The right side is the alternating harmonic series and the left side simplifies to {eq}ln(2) {/eq}.

- So, the alternating harmonic series converges to {eq}ln(2) {/eq}.

The **harmonic series** is the series: {eq}\sum_{n=1}^{\infty} \frac{1}{n} {/eq}. The harmonic series is a useful series that arises in mathematics and it has special usefulness in music since the harmonic series forms a basis of all tone systems. Now, a **harmonic number** is a number that can be represented as a partial sum of the harmonic series. That is if the harmonic series is taken only to a finite number of values, whatever number this sums to is called a harmonic number. Unfortunately, there is no easy way to sum a harmonic series. The only way to effectively sum a partial harmonic series is to add the terms that arise in the summation.

Finally, the series can either converge or diverge. A series is **convergent** if its sequence representation is convergent and its nth partial sum converges to a real number. If either of these conditions fail, then the series is said to be **divergent**. If a series is divergent, then it either has no real number output or it goes to infinity rather than a real number. It can be shown that the harmonic series is divergent using classical methods or methods that arise in calculus, such as the integral test.

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Frequently Asked Questions

The harmonic series is the series starting at 1 and going to infinity of 1/n. It looks like 1 + 1/2 + 1/3 + 1/4 +.... It is the series of rational numbers whose numerators are one and whose denominators are integers increasing by one each time.

The harmonic series does not converge absolutely. Simply take the absolute value of 1/n and the output is still 1/n. Therefore, the absolute value of the harmonic series is the harmonic series which diverges.

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