Does $f(n) = \Theta ( g(n ) )$ implies $f(n) \sim \gamma g(n)$ for $n \rightarrow \infty$ Saying that $f_n = \Theta(g_n)$ means there are two real constants $c_1,c_2$ and two integers $n_1,n_2$ such that

$$
c_1 g_n \leq f_n \leq c_2 g_n 
$$

for all $n \geq n_0 = \max(n_1,n_2)$, I have the feeling that because of this there's a constant $c_1 &lt; \gamma &lt; c_2$ such that
$$
f_n \sim \gamma g_n.
$$

You can assume also that both $g(n)$ and $f(n)$ are strictly increasing sequences and that specifically

$$
c_1 g_n &lt; f_n &lt; c_2 g_n 
$$

for $n \geq n_0$
My attempt to formalize was the following, for all $n$ I define the function

$$
c(\alpha)=c_1+(c_2-c_1)\alpha
$$
where $\alpha\in[0,1]$. And what I need to prove is that there's an $\alpha \in (0,1)$ such that

$$
f_n \sim c(\alpha)g_n
$$

as $n$ tends to $+\infty$. This implies I have to show

$$
\lim_{n\rightarrow+\infty} \frac{f_n}{c(\alpha)g_n}=1
$$

Or equivalently

$$
f_n = c(\alpha)g_n + o(g_n)
$$

for some $\alpha \in (0,1)$, but I don't actually know what to do... any clue? All I managed to prove is that $g_n$ can be multiplied by another sequence such that they're equivalent. But I can't prove that such sequence would converge.
No, it does not. 

Counterexample 1: Here is a counterexample for functions:
$$
g(x)=3, \qquad f(x)=3+\sin x.
$$
and similarly for sequences:
$$
g_n=3, \qquad f_n=3+(-1)^n.
$$

Counterexample 2: Here is a more interesting counterexample for monotonic functions:
$$
g(x) = 3x, \qquad f(x)=3x+x\sin \log x, \qquad x\ge1;
$$
so
$$
f'(x) = (3 x + x \sin(\log x))' = \sin \log x + \cos \log x + 3 &gt; 0.
$$

Similarly for monotonic sequences:
$$
g_n = 3n, \qquad f_n=3n+n\sin \log n, \qquad n\ge1.
$$
No, the claim is wrong because of the hidden constants.

Let $f(n)=2g(n)$. Then $f(n)=\Theta(g(n))$ but $f(n)\sim2g(n)\nsim g(n)$.