harmonic series
HarmonicSeries
2009-05-28 17:29:12
2009-05-29 08:35:31
Topic
Taylor series
no parent
necessary condition of convergence
Euler constant
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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}
$$\lim_{k\to\infty}a_n \;=\; 0$$
for the series \,$a_1+a_2+a_3+\ldots$ of real or complex terms:
$$\lim_{k\to\infty}\frac{1}{k} \;=\; 0$$
Nevertheless, the harmonic series diverges.\, It is seen if we first group the terms with parentheses:
$$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)
+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)
+\left(\frac{1}{9}+\frac{1}{10}+\ldots+\frac{1}{16}\right)+\ldots$$
Here, each parenthetic sum contains a number of terms twice as many as the preceding one.\, The sum in the first parentheses is greater than\, $2\cdot\frac{1}{4} = \frac{1}{2}$,\, the sum in the second parentheses is greater than\, $4\cdot\frac{1}{8} = \frac{1}{2}$;\, thus one sees that the sum in all parentheses is greater than $\frac{1}{2}$.\, Consequently, the partial sum of $n$ first terms exceeds any given real number, when $n$ is sufficiently big.\\
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
$n \to \infty$ (e.g. $\ln{1\,000\,000\,000} \,\approx\, 20.7$).\, The increasing 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 little positive number
$$\gamma \;=\; 0.5772156649...$$
which is called the \emph{Euler constant} or \emph{Euler--Mascheroni constant}.
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