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. Despite the terms approaching zero, the series diverges, meaning the sum grows without bound. In particular, since the harmonic series is a series with positive terms, 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:
\[S = \underset{n \rightarrow + \infty}{lim} s_{n} = \underset{n \rightarrow + \infty}{lim} \sum_{k = 1}^{n} \frac{1}{k}\]At first glance, one might think that as $n \rightarrow \infty$, 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, as shown below:
\[\sum_{k = 1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + ( \frac{1}{3} + \frac{1}{4} ) + ( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} ) + \hdots\]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 $a$:
\[\sum_{k = 1}^{\infty} \frac{1}{k^{a}}\]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:
\[\underset{k \rightarrow \infty}{lim} a_{k} = 0.\]If this condition is not satisfied, that is, if $\underset{k \rightarrow \infty}{lim} a_{k} \neq 0$, 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:
\[\underset{k \rightarrow \infty}{lim} \frac{1}{k^{0}} = \underset{k \rightarrow \infty}{lim} 1 = 1 \neq 0.\]Hence, the series diverges because its general term does not tend to zero.
When $a > 1$, we can apply the integral test to determine the convergence of the series. Consider the corresponding improper integral, we have:
\[\int_{1}^{\infty} \frac{1}{x^{a}} d x\]Since $a > 1$, we have:
\[\int_{1}^{\infty} \frac{1}{x^{a}} d x = \underset{t \rightarrow \infty}{lim} \int_{1}^{t} x^{- a} d x\]By evaluating the integral at the endpoints, we obtain:
\[\underset{t \rightarrow \infty}{lim} ([ \frac{x^{1 - a}}{1 - a} ])_{1}^{t} = \underset{t \rightarrow \infty}{lim} ( \frac{t^{1 - a}}{1 - a} - \frac{1}{1 - a} )\]Because $a > 1$, the exponent $1 - a < 0$, so $t^{1 - a} \rightarrow 0$. Therefore:
\[\int_{1}^{\infty} \frac{1}{x^{a}} d x = \frac{1}{a - 1}\]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 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, and decreasing for $x \geq 2$. We then evaluate the improper integral:
\[\int_{2}^{\infty} \frac{1}{x ( log x )^{\alpha}} d x\]Using the substitution $u = log x$, we get:
\[\int_{2}^{\infty} \frac{1}{x ( log x )^{\alpha}} d x = \int_{log 2}^{\infty} \frac{1}{u^{\alpha}} d u\]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.