# 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.

## 1 Answer

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

• Marvelous, thanks! – Boaz Yakubov Nov 25 '18 at 18:10