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 $c_1 < \gamma < c_2$ 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
$$
c_1 g_n < f_n < c_2 g_n 
$$
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.
 A: 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
$$
f'(x) = (3 x + x \sin(\log x))' = \sin \log x + \cos \log x + 3 > 0.
$$
Similarly for monotonic sequences:
$$
g_n = 3n, \qquad f_n=3n+n\sin \log n, \qquad n\ge1.
$$
A: Let $f(n)=a^n$ for some $a$. Let $$g(n)=\begin{cases}c_1f(n)&n\text{ odd}\\c_2f(n)&n\text{ even}\end{cases}$$
For $g$ to be monotonic, you need $c_1a^n>c_2a^{n-1}$ and $c_2a^n>c_1a^{n-1}$. This means that $c_2a>c_1$ and $c_1a>c_2$, or $\frac{1}{a}<\frac{c_1}{c_2}<a.$
So, for example, we can take $a=e, c_1=1,c_2=2$.
A: Intuition for why it's not true:
$f=\Theta(g)$ says that $f$ is never "that far" from $g$ in the sense of their ratio falling in $(c_1,c_2)$ for some $c_1,c_2$. However, this says nothing about how well behaved that ratio is, just that it falls within certain bounds. It should be obvious that there are non-convergent sequences that only take on values in $(c_1,c_2)$. Even if you require $f,g$ to be monotone, that doesn't guarantee that their ratio is monotone if they have wildly fluctuating higher-order derivatives so that $f$ gains on and falls behind $g$ (in the sense of their ratio decreasing or increasing) at an erratic rate as $x$ increases.
