What is a Harmonic Series?

Harmonic Number

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.

Harmonic Series Formula

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.

Harmonic Series Diverges

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:

Fig. 1: Integral Test for Harmonic Series