harmonic seriesHarmonicSeries2009-05-28 17:29:122026-09-13 17:12:44Topicadded related: SimpleHarmonicOscillatornecessary condition of convergenceEuler constant% Object 784: Harmonic Series
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\usepackage{amsmath}
\usepackage{graphicx}The \emph{harmonic series}
\[
\sum_{k=1}^\infty \frac{1}{k}
=
1+\frac{1}{2}+\frac{1}{3}+\ldots
\]
satisfies the \emph{necessary condition of convergence}
\[
\lim_{k\to\infty} a_k = 0
\]
for a series
\[
a_1+a_2+a_3+\ldots
\]
of real or complex terms. In particular,
\[
\lim_{k\to\infty}\frac{1}{k}=0.
\]
Nevertheless, the harmonic series diverges. This can be seen by grouping
the terms as
\[
1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)
+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)
+\left(\frac{1}{9}+\frac{1}{10}+\ldots+\frac{1}{16}\right)+\ldots
\]
Here, each parenthesized sum contains twice as many terms as the preceding
one. The sum in the first parentheses is greater than
\[
2\cdot\frac{1}{4}=\frac{1}{2},
\]
and the sum in the second parentheses is greater than
\[
4\cdot\frac{1}{8}=\frac{1}{2}.
\]
Continuing in this way, one sees that every parenthesized block is greater
than $\frac{1}{2}$. Consequently, the partial sum of the first $n$ terms
eventually exceeds any given real number, and therefore the harmonic series
diverges.
The \PMlinkescapetext{divergence} of the harmonic series is very slow,
though. Its \PMlinkescapetext{speed} may be illustrated by considering the
difference
\[
\sum_{k=1}^{n-1}\frac{1}{k}-\int_1^n\frac{dx}{x}
=
\sum_{k=1}^{n-1}\frac{1}{k}-\ln n
\]
(see the \PMlinkescapetext{diagram}). We know that $\ln n$ increases very
slowly as $n\to\infty$ (for example,
$\ln(1\,000\,000\,000)\approx 20.7$). The increase of the partial sum
$\sum_{k=1}^{n-1}\frac{1}{k}$ is about the same, since the limit
\[
\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\ln n\right)=\gamma
\]
is a small positive number,
\[
\gamma = 0.5772156649\ldots,
\]
called the \emph{Euler constant} or \emph{Euler--Mascheroni constant}.
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