The $n^{th}$ harmonic number $H_n$ is defined as
$$H_n = \sum_{k=1}^{n} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n}$$
It is known that $H_n$ is never an integer except for $n=1$.
The problem in this post is to show that we can get arbitrarily close.
i.e.
Show that there is an ascending sequence of integers $n_1 \lt n_2 \lt n_3 \lt \dots $ such that
$$ \lim_{k \to \infty} \{H_{n_k}\} = 0$$
where $\{x\}$ is the fractional part of x. Eg, $\{H_2\} = \frac{1}{2}, \{H_3\} = \frac{5}{6}$
In general it seems like a difficult problem to estimate the fractional parts of $H_n$. So if you got here by googling for information on that, sorry, this blog post won't be of much help.
[Solution]
$$H_n = \sum_{k=1}^{n} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n}$$
It is known that $H_n$ is never an integer except for $n=1$.
The problem in this post is to show that we can get arbitrarily close.
i.e.
Show that there is an ascending sequence of integers $n_1 \lt n_2 \lt n_3 \lt \dots $ such that
$$ \lim_{k \to \infty} \{H_{n_k}\} = 0$$
where $\{x\}$ is the fractional part of x. Eg, $\{H_2\} = \frac{1}{2}, \{H_3\} = \frac{5}{6}$
In general it seems like a difficult problem to estimate the fractional parts of $H_n$. So if you got here by googling for information on that, sorry, this blog post won't be of much help.
[Solution]
