Asymptotic equivalence and sums

Question

Suppose that for every $k \geq 1$:

$$ f_k (x) \sim c_k g(x), \quad x \to \infty,$$

where $f_k(x)$ and $g(x)$ are some positive functions and the $c_k$ nonnegative constants. My question is whether for every $m \geq 1$: $$ \sum_{k=1}^m f_k (x) \sim \sum_{k=1}^m c_k g(x), \quad x \to \infty. \label{1}\tag{1} $$ I would think so, because when considering the limit of the quotient of both sides, one could just divide by $g(x)$. The harder question is whether then even $$ \sum_{k=1}^\infty f_k(x) \sim \sum_{k=1}^\infty c_k g(x), \quad x \to \infty. \label{2}\tag{2}$$ Here, I am not sure anymore that this is true. Does this follow from \eqref{1}?

Note that $f(x) \sim g(x)$ if $f(x)/ g(x) \to 1$ as $x \to \infty$.

Answer 1

No, it is false in general. Equivalence is compatible with products, quotients or exponentiation to a fixed power, but not with addition in general.

Counterexample:

$1+x\sim_0 1$, $-1\sim -1$, but $1+x -1\not \sim_0 1-1=0$.

Answer 2

For the simplicity I assume $x\in (0,\infty)$. The equality \eqref{2} can fail, for instance, when $f_k(x)=1$, $g_k(x)=1+2^k/x$, and $c_k=1/2^k$ for each $k$ and $x$. In this case the left-hand side of \eqref{2} equals $1$, whereas the right-hand side is unbounded.