# Harmonic Series
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/harmonic-series/
Fetched from algebrica.org post 16071; source modified 2026-04-12T20:23:33.
## What is a harmonic series
The **harmonic series** is defined as the infinite sum:
\\[
\\sum\_{k=1}^{\\infty} \\frac{1}{k}
\\]
where each term is the reciprocal of a [natural number](../natural-numbers.md). Despite the terms approaching zero, the [series](../series.md) diverges, meaning the sum grows without bound. In particular, since the harmonic series is a series with [positive terms](../series-with-positive-terms.md), it can be shown that it diverges to positive infinity. In fact, since the series has positive terms, the limit of the sequence of its partial sums exists:
\\[
S = \\lim\_{n \\to +\\infty} s\_n = \\lim\_{n \\to +\\infty} \\sum\_{k=1}^{n} \\frac{1}{k}
\\]
At first glance, one might think that as \\( n \\to \\infty \\), the terms of the harmonic series tend to zero and therefore the series itself might converge. However, this is a logical error: although the terms \\( \\frac{1}{n} \\) do approach zero, they do not do so fast enough for the series to converge.
##### Here is the graph of the partial sums of the harmonic series up to \\(n = 100.\\) As can be seen, the curve rises slowly yet unceasingly, confirming that the series is divergent.
* * *
There are several ways to demonstrate that the harmonic series diverges. One approach involves using [partial sums](../series.md), as shown below:
\\[
\\sum\_{k=1}^{\\infty} \\frac{1}{k} = 1 + \\frac{1}{2} + \\left( \\frac{1}{3} + \\frac{1}{4} \\right) + \\left( \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\right) + \\cdots
\\]
Each group contains twice as many terms as the previous one. By observing that each group adds at least \\( \\frac{1}{2} \\) to the sum, we conclude that the overall series grows without bound:
\\[
\\sum\_{k=1}^{\\infty} \\frac{1}{k} = \\infty
\\]
Hence, the harmonic series diverges.
Knowing how the harmonic series and its variants behave is useful, since the comparison test often lets us relate a complex series to a harmonic one to check whether it converges.
## Generalized harmonic series (p-series)
Let’s consider a generalized version of the harmonic series, where the denominator is raised to a [power](../powers.md) \\( a \\):
\\[
\\sum\_{k=1}^{\\infty} \\frac{1}{k^a}
\\]
This series, known as **generalized harmonic series**, has an additional feature compared to the standard harmonic series: its convergence or divergence depends on the value of the exponent \\( a \\). We have:
- If \\( a > 1 \\), the series converges.
- If \\( a \\leq 1 \\), the series diverges.
* * *
A necessary condition for the convergence of a series \\( \\sum a\_k \\) is that the general term tends to zero:
\\[
\\lim\_{k \\to \\infty} a\_k = 0.
\\]
If this condition is not satisfied, that is, if \\( \\lim\_{k \\to \\infty} a\_k \\neq 0 \\), then the series diverges. In the case of the generalized harmonic series with exponent \\( a \\leq 1 \\), this condition fails. For example, when \\( a = 0 \\), the terms become constant \\( a\_k = 1 \\), and:
\\[
\\lim\_{k \\to \\infty} \\frac{1}{k^0} = \\lim\_{k \\to \\infty} 1 = 1 \\neq 0.
\\]
Hence, the series diverges because its general term does not tend to zero.
* * *
When \\( a > 1 \\), we can apply the [integral test](../integral-test-for-series-convergence.md) to determine the convergence of the series. Consider the corresponding improper integral, we have:
\\[
\\int\_1^{\\infty} \\frac{1}{x^a} \\, dx
\\]
Since \\( a > 1 \\), we have:
\\[
\\int\_1^{\\infty} \\frac{1}{x^a} \\, dx = \\lim\_{t \\to \\infty} \\int\_1^t x^{-a} \\, dx
\\]
By evaluating the integral at the endpoints, we obtain:
\\[
\\lim\_{t \\to \\infty} \\left\[ \\frac{x^{1-a}}{1 - a} \\right\]\_1^t = \\lim\_{t \\to \\infty} \\left( \\frac{t^{1 - a}}{1 - a} - \\frac{1}{1 - a} \\right)
\\]
Because \\( a > 1 \\), the exponent \\( 1 - a < 0 \\), so \\( t^{1 - a} \\to 0 \\). Therefore:
\\[
\\int\_1^{\\infty} \\frac{1}{x^a} \\, dx = \\frac{1}{a - 1}
\\]
which is finite. Hence the series converges for all \\( a > 1 \\).
## Logarithmically modified harmonic series
Finally, let us consider the following series:
\\[
\\sum\_{k=2}^{\\infty} \\frac{1}{k (\\log^\\alpha k)}
\\]
This is a so-called **logarithmically modified** series, where the presence of the [logarithmic](../logarithms.md) term affects the rate at which the series converges or diverges. The convergence of the series depends on the exponent \\( \\alpha \\) in the logarithmic term:
- If \\( \\alpha > 1 \\), the series converges.
- If \\( \\alpha \\leq 1 \\), the series diverges.
The summation starts at \\( k = 2 \\) to avoid the singularities at \\( k = 0 \\) (where \\( \\log 0 \\) is undefined) and \\( k = 1 \\) (where \\( \\log 1 = 0 \\), causing division by zero).
* * *
To demonstrate the convergence, we apply the improper integral test. We consider the function:
\\[
f(x) = \\frac{1}{x (\\log x)^\\alpha}
\\]
which is positive, [continuous](../continuous-functions.md), and [decreasing](../increasing-and-decreasing-functions.md) for \\( x \\geq 2 \\). We then evaluate the improper integral:
\\[
\\int\_2^{\\infty} \\frac{1}{x (\\log x)^\\alpha} \\, dx
\\]
Using the substitution \\( u = \\log x \\), we get:
\\[
\\int\_2^{\\infty} \\frac{1}{x (\\log x)^\\alpha} \\, dx = \\int\_{\\log 2}^{\\infty} \\frac{1}{u^\\alpha} \\, du
\\]
This integral converges if and only if \\( \\alpha > 1 \\). Therefore, by the integral test, the series converges if and only if \\( \\alpha > 1 \\).
## Example
Let us determine the nature of the following series:
\\[
\\sum\_{n=2}^{\\infty} \\frac{1}{n \\sqrt{\\log n}}
\\]
* * *
We observe that this series has the general form:
\\[
\\sum\_{n=2}^{\\infty} \\frac{1}{n (\\log n)^\\alpha}
\\]
with \\( \\alpha = \\frac{1}{2} \\). This is known as a logarithmically modified harmonic series. According to known results, the series converges if and only if \\( \\alpha > 1 \\).
Since \\( \\alpha = \\frac{1}{2} < 1 \\), the series diverges.
We conclude that the series diverges by comparison with a divergent harmonic series modified by a logarithmic term.
