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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!

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    $\begingroup$ I suspect that when you write “series”, what you really mean is “sequence”. Am I right? $\endgroup$ – José Carlos Santos Apr 20 '19 at 21:16
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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 $n: p_1<n<q_1$ there exists $r_1$ such that $x_{r_1}\in U_{\varepsilon}(a)$

And so on...

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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 $a_{i - 1} < x < a_i < a_{i - 1} + \varepsilon$.

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