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

Question

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.

Answer 1

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. $\lim_j a_{n_j} = L$.