= Harmonic series
{title2=$\sum_{n=1}^{\infty}\frac1n$}
{wiki}
The <harmonic series> is the positive <series>
$$
1+\frac12+\frac13+\frac14+\cdots=\sum_{n=1}^{\infty}\frac1n.
$$
Its <partial sums> are the <harmonic numbers> $H_N$. Although the individual terms tend to zero, $H_N$ grows without bound, so the <harmonic series> is not a <convergent series>. This is a basic warning against using the size of one term as a bound on all the terms still to come.
The growth is slow enough to be visually deceptive. The <partial sums> at $N=10,100,1000,10000$ are approximately $2.92896825,5.18737752,7.48547086,9.78760604$. Adding ten times as many terms increases the accumulated total by roughly $\log10\approx2.30258509$, rather than multiplying the total by ten. The <telescoping series> $\sum1/[n(n+1)]$ has similarly vanishing terms but approaches the finite sum one.
\Image[/harmonic-versus-telescoping-sums.png]
{title=Small terms can accumulate without bound or approach a finite sum}
{description=The left panel compares the positive terms $1/n$ and $1/[n(n+1)]$ on <logarithmic scales>. The right panel compares their <partial sums> against the number of terms on a logarithmic horizontal axis. The dashed curve is $\log N$, a lower comparison for $H_N$; the horizontal line is the finite sum one of the <telescoping series>.}
A short block argument proves the divergence without trusting a finite plot. For each <integer> $k\geq0$, the block from $2^k+1$ through $2^{k+1}$ contains $2^k$ terms, each at least $1/2^{k+1}$. Therefore
$$
\sum_{n=2^k+1}^{2^{k+1}}\frac1n\geq\frac12,\qquad
H_{2^m}\geq1+\frac m2.
$$
Every doubling supplies at least another half, and arbitrarily many doublings are possible. A collection of tiny terms can still have a substantial combined contribution. See https://www.jirka.org/ra/html/sec_series.html[Jiří Lebl, Basic Analysis I, Example 2.5.11] for this grouping proof.
<Integral comparison> sharpens the growth estimate. Here $\log$ denotes the <natural logarithm>. Since $1/x$ is a <strictly decreasing function> for positive $x$, comparison of rectangles with the <integral> gives
$$
\log(N+1)\leq H_N\leq1+\log N.
$$
Thus the divergence is logarithmic. The finer difference $H_N-\log N$ tends to the <Euler--Mascheroni constant>, approximately $0.57721566$; the <harmonic number> expansion supplies higher-order corrections. This turns the counterexample into a usable quantitative estimate of accumulated reciprocal contributions.
Euler's \i[De progressionibus harmonicis observationes], written in 1734 and published in 1740, studies these sums beyond the yes-or-no question of convergence. Sections 2–3 compare long blocks to show continued growth, and later sections compare the sums with <logarithms> and calculate the constant offset. The https://scholarlycommons.pacific.edu/euler-works/43/[Euler Archive record, E43]{ref} identifies the dates and original publication; https://download.uni-mainz.de/mathematik/Algebraische%20Geometrie/Euler-Kreis%20Mainz/E43.pdf[Alexander Aycock's translation of Euler's original paper]{ref} makes the calculations accessible. This is not a claim that Euler first discovered divergence: the paper's interest includes measuring how slowly a divergent sum grows.
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