# A homework question about partial limits

I could really use some help figuring out this question.

The question: $${a_n}$$ is a series so that $$\lim_{n\to\infty} (a_{n+1} - a_n) = 0$$. Prove that its group of partial limits is the closed interval $$[\liminf \, a_n, \limsup\, a_n]$$.

Solution attempt:

One direction of the proof seems to be trivial - all the partial limits of a series are between the liminf and the limsup. But I have no idea how to prove the other direction - given an $$L$$ in this interval, what makes it a partial limit?

A hint (or a solution) would be appriciated. Thank you in advance!

• I suspect that when you write “series”, what you really mean is “sequence”. Am I right? – José Carlos Santos Apr 20 '19 at 21:16

Let $${liminf \{a_n\}}=A, limsup \{a_n\}=B$$, we take $$L \in [A; B]$$ and we want to show that $$L$$ is a partial limit. It's enough to show that $$\forall \varepsilon>0$$ we have $$U_{\varepsilon}(L)$$ contains infinitely many elemets of sequence $$\{a_n\}$$.
We may assume that $$\varepsilon>0$$ is small enough, so that $$U_{\varepsilon}(L)$$,$$U_{\varepsilon}(A)$$ and $$U_{\varepsilon}(B)$$ do not intersect. From the condition $$\lim_{n\to\infty} (a_{n+1} - a_n) = 0$$ we can take $$N=N(\varepsilon)$$ such that $$|a_{n+1} - a_n|<2\varepsilon$$ when $$n>N$$
As soon as $$A$$ is a partial limit, there exists $$p_1>N$$ such that $$x_{p_1}\in$$ $$U_{\varepsilon}(A)$$.For the same reason there exists $$q_1>p_1$$ such that $$x_{q_1}\in$$ $$U_{\varepsilon}(B)$$. But we know that $$|a_{n+1} - a_n|<2\varepsilon$$ when $$n>N$$, so among $$n: p_1 there exists $$r_1$$ such that $$x_{r_1}\in U_{\varepsilon}(a)$$
Take $$x \in [\liminf a_n, \limsup a_n]$$, $$\varepsilon > 0$$ and $$N$$, we need to find $$k > N$$ such that $$a_k \in U_\varepsilon(x)$$.
Let $$M$$ be such that $$\forall n > M: |a_{n + 1} - a_n| < \varepsilon$$. Choose $$p > q > M$$ such that $$a_p \in U_\varepsilon(\liminf a_n)$$ and $$a_q \in U_\varepsilon(\limsup a_n)$$. If $$x < a_p$$ then take $$k = p$$, if $$x > a_q$$ then take $$k = q$$. If $$a_p < x < a_q$$ then find minimal $$i \in \{p, p + 1, \ldots, q\}$$ such that $$a_i > x$$ and choose $$k = i$$: we have $$a_{i - 1} < x < a_i < a_{i - 1} + \varepsilon$$.