Does the set $\{1 + 1/2 + 1/3 + \cdots + 1/n : n ∈ \mathbb{N}\}$ have any limit points? Does the set $A = \{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} : n\in \mathbb{N}\}$ have any limit points?
My intuition is that it does not.
No element in $A$ is a limit point, since for any element $x$ in $A$ there is an $\epsilon$-neighborhood of $x$ intersecting $A$ only at $x$. It also seems that I can prove using the density of $\mathbb{Q}$ in $\mathbb{R}$ that any number not in $A$ has some $\epsilon$-neighborhood that intersects $A$ nowhere.
Yet, I doubt that this method will work, because I could do the same thing for             $\{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} : n\in \mathbb{N}\}$, even though this set definitely has a limit point, namely Euler's Sum.
 A: Yes, your intuition is right but you're going to have to be more careful in arguing that no real not in $A$ is a limit point of $A$. In particular you're going to have to use a fact about the harmonic series which is not true for the sum of the reciprocals of squares - namely, that it diverges! Suppose $\lambda$ were a limit point of $A$. Then since the harmonic series diverges, we have $\lambda<a$ for some $a\in A$. Do you see how to go from here?

In fact, the following is true (and a good exercise):

Suppose $\{a_i: i\in\mathbb{N}\}$ is a set of positive numbers. Then $A=\{\sum_{i=1}^na_i: n\in\mathbb{N}\}$ has at most one limit point, and that limit point (if it exists) is $\sum_{i=1}^\infty a_i$.

A: The series $\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges, this means that for every $l\in\mathbb R$ there exists $N$ such that $x_n> l+1$.
So if we take $l\in \mathbb R $ and $\epsilon<1$ then there is only a finite number of elements of $a\in A$ for which we could have $a\in (l-\epsilon,l+\epsilon)$. Taking $\epsilon$ smaller than the minimum of all the distances between those sets of $A$ (different from $l$) gives us an open ball $B$ around $l$ such that $(B\setminus l ) \cap A=\varnothing$, so no poitn is a limit point of $A$.
A: Consider the partial sums $S_k = \sum\limits_{n=1}^k \frac{1}{n}$. Then $S_1 < S_2 < \cdots$ and $A = \{S_n : n \in \mathbb N\}$. Thus $(S_{n-1}, S_{n+1})$ is an open set containing $S_n$ but not containing any point of $A$ other than $S_n$. So, no point of $A$ is a limit point of $A$. This works for any positive series. In fact, the limit point of the set $A$ exists if and only if the series converges, and, in that case, the sum (which is also equal to the supremum of $A$) is the limit point of $A$.
Added later: If $a < S_1$, then $(-\infty,S_1)$ is an open set containing $a$ but not containing any element of $A$. If $S_n < a < S_{n+1}$, then $(S_n < a < S_{n+1})$ is an open set containing $a$ but not containing any element of $A$. Since $\sup A = \infty$, the proof is complete. However, if $\sup A < \infty$, then it is the limt point of $A$.
