Proving the unboundedness of a particular real-valued sequence. 
Proposition Suppose that $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ \mathbb{R} $ such that
  $$
\forall m,n \in \mathbb{N}: \quad m > n ~~ \Longrightarrow ~~ |x_{m} - x_{n}| > \frac{1}{n}.
$$
  Then $ (x_{n})_{n \in \mathbb{N}} $ is unbounded.

I can somehow show that $ \displaystyle \text{Diam} \left( \{ x_{n} \}_{n=1}^{N} \right) > \sum_{n=1}^{N} \frac{1}{n} $ for all $ N \in \mathbb{N} $, but the details are messy. Might there be a slicker proof? Thank you very much!
 A: For every $n$, let $I_n=(x_n-\frac1{2n},x_n+\frac1{2n})$. Assume that for some $n\lt m$, there exists $x$ in $I_n\cap I_m$. Then $|x_n-x_m|\leqslant|x-x_n|+|x-x_m|\lt\frac1{2n}+\frac1{2m}\lt\frac1n$, which is absurd. Hence $I_n\cap I_m$ is empty for every $n\ne m$.
In particular, the set $J=\bigcup\limits_nI_n$ has Lebesgue measure $\sum\limits_n\frac1n$, which is infinite, thus $J$ is not bounded. For every $M\gt0$ there exists $x$ in $J$ such that $|x|\gt M$. By definition of $J$, this implies that there exists $n$ such that $|x_n|\geqslant|x|-|x-x_n|\gt M-1$, hence the sequence $(x_n)_n$ is not bounded.
Note that one can replace $\frac1n$ by some positive $a_n$, as long as the series $\sum\limits_na_n$ diverges.
A: Here is the solution that I have been working on, which avoids measure theory. I have managed to simplify my original argument, so it is now worthy of being posted! :)

Let $ N \in \mathbb{N} $.

    
*
    
*Observe that the terms of $ (x_{n})_{n \in \mathbb{N}} $ are distinct.
    
*Find a permutation $ \sigma: \{ 1,\ldots,N + 1 \} \to \{ 1,\ldots,N + 1 \} $ that makes $ (x_{\sigma(n)})_{n=1}^{N+1} $ a strictly increasing sequence.
    
*Let $ n \in \{ 1,\ldots,N \} $.
    
    
*
    
*If $ \sigma(n + 1) > \sigma(n) $, then the hypothesis yields $ x_{\sigma(n + 1)} - x_{\sigma(n)} > \dfrac{1}{\sigma(n)} $.
    
*If $ \sigma(n + 1) < \sigma(n) $, then the hypothesis yields $ x_{\sigma(n + 1)} - x_{\sigma(n)} > \dfrac{1}{\sigma(n + 1)} $. Hence, as $ \dfrac{1}{\sigma(n + 1)} > \dfrac{1}{\sigma(n)} $, we also obtain $ x_{\sigma(n + 1)} - x_{\sigma(n)} > \dfrac{1}{\sigma(n)} $.
    
    
*We thus have the following:
    \begin{align}
      \sum_{n=1}^{N} \left[ x_{\sigma(n + 1)} - x_{\sigma(n)} \right]
&>    \sum_{n=1}^{N} \frac{1}{\sigma(n)}. \quad
      (\text{Telescoping sum on the left-hand side.}) \\
      x_{\sigma(N + 1)} - x_{\sigma(1)}
&>    \sum_{n=1}^{N} \frac{1}{\sigma(n)} \quad
      (\text{After evaluating the telescoping sum.}) \\
&=    \left[ \sum_{n=1}^{N+1} \frac{1}{\sigma(n)} \right] -
      \frac{1}{\sigma(N + 1)} \quad (\text{Adding and subtracting.}) \\
&=    \left[ \sum_{n=1}^{N+1} \frac{1}{n} \right] - \frac{1}{\sigma(N + 1)}
      \quad (\text{As $ \sigma $ is a permutation.}) \\
&\geq \left[ \sum_{n=1}^{N+1} \frac{1}{n} \right] - 1 \quad
      (\text{As $ \sigma(N + 1) \geq 1 $.}) \\
&=    \sum_{n=2}^{N+1} \frac{1}{n}.
\end{align}
    
*Hence,
    $$
  \text{Diam} \left( \{ x_{n} \}_{n=1}^{N+1} \right)
= \left( \max_{1 \leq n \leq N + 1} x_{n} \right) -
  \left( \min_{1 \leq n \leq N + 1} x_{n} \right)
> \sum_{n=2}^{N+1} \frac{1}{n}.
$$

As $ N \in \mathbb{N} $ is arbitrary, it follows from the divergence of the Harmonic Series that $ \text{Diam} \left( \{ x_{n} \}_{n=1}^{\infty} \right) = \infty $.

Conclusion: $ (x_{n})_{n=1}^{\infty} $ is an unbounded sequence in $ \mathbb{R} $.
