If $\lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) =[\lim\inf(a_n),\lim\sup(a_n)]$ 
If $\{a_n\}$ is a series, and $\lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) = [\lim\inf(a_n),\lim\sup(a_n)]$

Basically, I wish to prove that the set of all partial limits of the series $a_n$ (as $n\to\infty$) contains all values between its maximal limit ($\lim\sup$) and minimal limit ($\lim\inf$)
I already know that $\lim\sup$ and $\lim\inf$ are limits of $a_n$ by definition, and so it is a closed set $[\lim\inf(a_n),\lim\sup(a_n)]$.
Since $\lim\sup(a_n) = \sup(P(a_n))$ and $\lim\inf(a_n) = \inf(P(a_n))$ then $P(a_n)$ is included in $[\lim\inf(a_n),\lim\sup(a_n)]$.
Now if I were to take any $\lim\inf(a_n) < L < \lim\sup(a_n)$, I need to prove that it is a partial limit.
 A: Let $\liminf_n a_n < L < \limsup_n a_n$, and let $k_0\in\mathbb{Z}^+$ be such that
$$
\liminf_n a_n < L-\frac{1}{k_0}\,,
\qquad
L + \frac{1}{k_0} < \limsup_n a_n.
$$
By assumption, there exists an index $N_0$ such that
$$
(1)\qquad
|a_{n+1}-a_n| < \frac{1}{k_0}
\qquad \forall n\geq N_0.
$$
Moreover, we can find indices $j_0, m_0 \geq N_0$ such that
$$
a_{j_0} < L - \frac{1}{k_0},
\qquad
L + \frac{1}{k_0} < a_{m_0}.
$$
Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
$n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < \frac{1}{k_0}$.
As a second step, let $N_1 > \max\{N_0, j_0, m_0\}$ be such that
$$
(2)\qquad
|a_{n+1}-a_n| < \frac{1}{k_0+1}
\qquad \forall n\geq N_1.
$$
Moreover, we can find indices $j_1, m_1 \geq N_1$ such that
$$
a_{j_1} < L - \frac{1}{k_0+1},
\qquad
L + \frac{1}{k_0+1} < a_{m_1}.
$$
Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
$n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < \frac{1}{k_0+1}$.
We can in this way construct by induction a sequence $n_0 < n_1 < \cdots$ such that
$$
|a_{n_j} - L| < \frac{1}{k_0+j}
\qquad\forall j,
$$
i.e. $\lim_j a_{n_j} = L$.
