# Prove that $\lim_{n \to \infty} a_n = -\infty \$ if $\ \ \forall n\geq m: \ a_n \leq a_m +2017$, and $-\infty$ is a substantial limit of $a_n$

Given a sequence $$\{a_n\}_{n=1}^{\infty}$$ such that

$$\forall n\geq m: \ a_n \leq a_m +2017$$,

and $$-\infty$$ is a substantial limit of $$a_n$$,

Prove that $$\lim_{n \to \infty} a_n = -\infty$$.

my attempt: It's clear that: $$\text{lim inf}_{n \to \infty} (a_n) = -\infty$$.

I'd like to prove that $$\text{lim inf}_{n \to \infty} (a_n) = -\infty$$, and that will prove the claim.

there exist a sub-sequence of $$a_n$$,

let it be $$a_{n_{k}}$$ such that $$lim_{k\to \infty} a_{n_{k}} = -\infty$$

then for every $$n \geq m: \ \$$ $$a_{n_{k}} \leq a_{m_{k}} + 2017$$

$$\exists M > 0 \text{ such that} \ \forall N\in \mathbb{N}: \exists n \geq m \geq N: M \leq |a_{n_{k}}| \leq |a_{m_{k}} + 2017|$$

I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.

Fix $$m$$ then $$\limsup_{n\to\infty } a_n \leq \limsup_{n\to\infty } (a_m +2017 )= a_m +2017$$ Now $$\limsup_{n\to\infty } a_n=\liminf_{m\to\infty } ( \limsup_{n\to\infty } a_n) \leq \liminf_{m\to\infty } ( a_m +2017)=2017+\liminf_{m\to\infty } a_m =-\infty$$
If $$\liminf_{k\to\infty} a_{n_k} = -\infty$$ then for every $$M$$ there is some $$K$$ such that $$a_{n_k} < M$$ for every $$k \geq K$$, and by the condition stated, $$a_n \leq a_{n_k} + 2017 < M + 2017$$ for every $$n\geq n_k$$. If we pick $$M = -N-2017$$ then $$a_n < -N$$ for every $$n\geq n_k$$, so the tail of the sequence can be made arbitrarily small for $$n$$ large enough, which is the definition of $$\lim_{n\to\infty} a_n = -\infty$$.