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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.

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Fix $m$ then $$\limsup_{n\to\infty } a_n \leq \limsup_{n\to\infty } (a_m +2017 )= a_m +2017$$ Now $$\limsup_{n\to\infty } a_n=\liminf_{m\to\infty } ( \limsup_{n\to\infty } a_n) \leq \liminf_{m\to\infty } ( a_m +2017)=2017+\liminf_{m\to\infty } a_m =-\infty$$

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

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