Harmonic Series

Definition

1.The harmonic series is defined as the infinite sum:

$$ \sum_{k=1}^{\infty} \frac{1}{k} $$

2.where each term is the reciprocal of a natural number. Despite the terms approaching zero, the series diverges, meaning the sum grows without bound. Because it is a series with positive terms, the limit of the sequence of its partial sums exists, and the series diverges to positive infinity:

$$ S = \lim_{n \to +\infty} s_n = \lim_{n \to +\infty} \sum_{k=1}^{n} \frac{1}{k} $$

3.One might think that, as $n \to \infty$, the terms tend to zero and so the series converges. This is a logical error, the terms $\frac{1}{n}$ do approach zero but not fast enough for the series to converge.

4.

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.

5.Several methods prove that the harmonic series diverges. One groups the terms of the partial sums:

$$ \sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \cdots $$

6.Each group contains twice as many terms as the previous one. By observing that each group adds at least $1/2$ to the sum, we conclude that the overall series grows without bound:

$$ \sum_{k=1}^{\infty} \frac{1}{k} = \infty $$

7.Hence, the harmonic series diverges. Knowing how the harmonic series and its variants behave is useful, since the comparison test and the asymptotic comparison test often let us relate a complex series to a harmonic one to check whether it converges.

Generalized harmonic series (p-series)

8.Raising the denominator to a power $a$ produces the series:

$$ \sum_{k=1}^{\infty} \frac{1}{k^a} $$

9.Unlike the standard harmonic series, its convergence depends on the exponent $a$:

11.A necessary condition for the convergence of a series $\sum a_k$ is that the general term tends to zero:

$$ \lim_{k \to \infty} a_k = 0. $$

12.If this condition is not satisfied, that is, if $\lim_{k \to \infty} a_k \neq 0$, then the series diverges. For 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:

$$ \lim_{k \to \infty} \frac{1}{k^0} = \lim_{k \to \infty} 1 = 1 \neq 0. $$

13.Hence, the series diverges because its general term does not tend to zero.

14.When $a > 1$, we apply the integral test to determine whether the series converges. We consider the corresponding improper integral:

$$ \int_1^{\infty} \frac{1}{x^a} \ dx $$

15.Since $a > 1$, we have:

$$ \int_1^{\infty} \frac{1}{x^a} \ dx = \lim_{t \to \infty} \int_1^t x^{-a} \ dx $$

16.By evaluating the integral at the endpoints, we obtain:

$$ \lim_{t \to \infty} \left[ \frac{x^{1-a}}{1 - a} \right]_1^t = \lim_{t \to \infty} \left( \frac{t^{1 - a}}{1 - a} - \frac{1}{1 - a} \right) $$

17.Because $a > 1$, the exponent $1 - a < 0$, so $t^{1 - a} \to 0$. Therefore:

$$ \int_1^{\infty} \frac{1}{x^a} \ dx = \frac{1}{a - 1} $$

18.which is finite. Hence the series converges for all $a > 1$.

Abel series

19.Weighting the generalized harmonic series with a logarithmic factor produces the Abel series, which depends on two exponents $p$ and $q$:

$$ \sum_{n=2}^{\infty} \frac{1}{n^p (\log n)^q} $$

20.The summation starts at $n = 2$ to avoid the index $n = 1$, where $\log 1 = 0$ causes division by zero. The power $n^p$ sets the dominant rate of decay, while the logarithmic factor refines the behavior only at the boundary. Its nature follows four cases:

22.When $p \neq 1$ the power $n^p$ dominates the logarithm, so the series follows the generalized harmonic series. For $p > 1$, choose an exponent $s$ with $1 < s < p$. Since $(\log n)^q$ grows slower than any positive power of $n$, eventually:

$$ \frac{1}{n^p (\log n)^q} < \frac{1}{n^s} $$

23.The series $\sum \frac{1}{n^s}$ converges because $s > 1$, so the comparison test gives convergence. For $p < 1$, choose $s$ with $p < s < 1$. Eventually the inequality reverses:

$$ \frac{1}{n^p (\log n)^q} > \frac{1}{n^s} $$

24.The series $\sum \frac{1}{n^s}$ diverges because $s < 1$, so the original series diverges as well.

25.The case $p = 1$ is the boundary, where the logarithm alone decides. We apply the integral test to the function:

$$ f(x) = \frac{1}{x (\log x)^q} $$

26.which is positive, continuous, and decreasing for $x \geq 2$. Using the substitution $u = \log x$, the improper integral becomes:

$$ \int_2^{\infty} \frac{1}{x (\log x)^q} \ dx = \int_{\log 2}^{\infty} \frac{1}{u^q} \ du $$

27.The last integral converges if and only if $q > 1$. By the integral test the series with $p = 1$ converges if and only if $q > 1$. As an application, we study the nature of the series:

$$ \sum_{n=2}^{\infty} \frac{1}{n \sqrt{\log n}} $$

28.This is an Abel series with $p = 1$ and $q = \frac{1}{2}$. Since $p = 1$ and $q = \frac{1}{2} \leq 1$, the series diverges.