Prove that the set of partial sums of the sequence $(1/n)_n$ is closed

Question

I am working on a problem out of Abbott's book Understanding Analysis.

The problem asks whether $\{1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}|n \in \mathbb{N}\}$ is open, closed or neither.

I've already shown that the set is open. I am reasonably sure that the set is closed, (my reasoning is that the set has no limit points), but I'm having a lot of trouble showing it.

I know that the sequence of partial sums of ${(\frac{1}{n})}$ is unbounded, and that's why the harmonic series diverges. However, there's no guarantee that some subsequence of the partial sums doesn't converge (if it did, I know I would have to show that the limit point is in the set).

Is this the right approach, and if so can I get a hint on how to continue? Or am I taking the wrong approach?

Thank you!

Answer 1

No subsequence of the sequence of partial sums will converge.

Write $H_n = 1 + \frac{1}{2} + \cdots +\frac{1}{n}$. If a subsequence converges, then it means that there exist $n_1 < n_2 < \cdots$, such that the limit $\lim_{i\rightarrow \infty}H_{n_i}$ exists.

In particular, the sequence $(H_{n_i})_i$ is bounded.

But the condition $n_1 < n_2 < \cdots$ means that any integer $M$ will be eventually smaller than some $n_j$. This will contradict the divergence of $(H_n)$.

Answer 2

If you mean open in $\Bbb{R}$ then it is closed since its complement is $$(-\infty,1) \cup \bigcup_{n=1}^{\infty}(1+\dotsb+\frac{1}{n},1+\dotsb+\frac{1}{n+1})$$ which is open.

The set is not open in $\Bbb{R}$ since for an arbitrary $x$ the set you cannot find an open interval contained in the set.

Answer 3

This is the right approach. Take any sequence $(u_n)$ of $H:= \{ 1 + ... + \frac{1}{n} | n\in \mathbb{N}\}$. Suppose that (u_n) converges. Thus it is bounded by some constant $M > 0$. Therefore for any $n\in \mathbb{N}$, $$u_n \in H\cap [0,M]$$

which is a finite set due to the divergence of the harmonic series. Since $u_n$ converges it must be constant after some $n_0$ and therfore its limit is in $H$.

We have shown that $H$ is closed.