The harmonic mean of a set of
positive numbers is the reciprocal of the arithmetic mean (ordinary
average) of the reciprocals of the numbers. Find $\lim_{n\to\infty}
(H_n/n)$, where $H_n$ is the harmonic mean of the $n$ positive integers
$n+1$, $n+2$, $n+3, \dots n+n$.
Solution. We
have
\lim_{n\to\infty} \frac{H_n}{n} =\lim_{n\to\infty}
\frac1{n}
\frac{n}{\frac1{n+1}+\frac1{n+2}+\cdots +\frac1{n+n}}.
It will ease things considerably if we look at the reciprocal of this:
\begin{eqnarray}
\lim_{n\to\infty}\frac{n}{H_n} &=& \lim_{n\to\infty} n \left(
\frac1{n} \right) \left(\frac1{n+1}+ \frac1{n+2} +\cdots + \frac1{n+n}
\right) \\
&=& \lim_{n\to\infty} \left(\frac1{n}\right)
\left(\frac1{1+1/n} + \frac1{1+2/n} + \cdots \frac1{1+n/n} \right) \\
&=& \lim_{n\to\infty}\left(\frac1{n}\right)\sum_{k=1}^n
\frac1{1+k/n} \\
&=& \int_0^1 \frac1{1+x}\,dx \\
&=& \ln |1+x| \Big]_0^1 = \ln 2.
\end{eqnarray}
So $\lim_{n\to\infty} \frac{H_n}{n} = \frac1{\ln 2}$.