How can we show that $\lim_{n \to +\infty} f(n)=+\infty$?

Question

We suppose that $\lim_{n \to +\infty} f(n)=+\infty$.

I want to prove that if $f(n)=O(g(n)), c \in \mathbb{R}$, then $f(n)+c=O(g(n))$ .

$f(n)=O(g(n))$

That means that $\exists c_1>0, n_2 \in \mathbb{N}$, such that $\forall n \geq n_2$:

$$f(n) \leq c_1 g(n)$$

$\lim_{n \to +\infty} f(n)=+\infty$

This means that $\forall \epsilon>0 \ \exists n_0$, such that $\forall n \geq n_0$, $$f(n)> M$$

How can we use this, in order to show that $f(n)+c=O(g(n))$ ?

Answer 1

Here is a (stronger) result (with an easier proof) which implies yours:

Assume that $\liminf |f|\gt0$, then, for every finite $c$, $f+c\in O(f)$.

To apply this to your setting, note that the hypothesis that $\lim f=+\infty$ implies that $\liminf |f|\gt0$, and that if $f+c\in O(f)$ and $f\in O(g)$ then $f+c\in O(g)$.

Now can you prove the result above? One should start with the hypothesis that $\liminf |f|\gt0$, hence $|f(n)|\geqslant a$ for some $a\gt0$, for every $n$ large enough, say, for every $n\geqslant N$, hence $|f(n)+c|\leqslant$ $____$ for every $n\geqslant N$, hence...