Does $f(x) \sim g(x)$ imply certain results about the sums of those functions?

Question

Suppose you have two functions $f(x)$ and $g(x)$ which are asymptotically equal, $f(x) \sim g(x)$. Suppose that $\sum f(x)$ diverges. Does this imply that $\sum g(x)$ diverges as well? If not, are there other conditions along with this which can imply $\sum g(x)$ diverges? Can the same be done if $\sum f(x)$ converges and $g(x)$ has no singularities? I am asking this because if this is true then it can be used to prove that there are infinitely many primes by showing that one function diverges and another converges.

Answer 1

If $f, g \ge 0$, then yes it does (by limit comparison test). Otherwise, not necessarily. For example, consider

$$ f(n) = \frac{(-1)^n}{\sqrt{n}}, \ g(n) = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} $$

Then $f(n) \sim g(n)$ but $\sum_{n =1}^{\infty} f(n)$ converges while $\sum_{n =1}^ {\infty} g(n)$ diverges.

Answer 2

Just a long comment on the converse of Robert Israel's answer. The converse of this is not immediately obvious i.e even if $f(x) >0, g(x) > 0$ and $\sum_{x \le n} f(x) \sim \sum_{x \le n} g(x)$ we cannot automatically conclude that. A very good example of this was the proof of the prime number theorem. It was already known that

$$ \sum_{n \le x}\dfrac{1}{n\log n} \sim \log \log x \text{, } \sum_{p_n \le x}\dfrac{1}{p_n} \sim \log \log x $$ but it took many more years to prove that indeed $p_n \sim n \log n$. This was one of the few discussions in MO where Terrence Tao and participated.

Why could Mertens not prove the prime number theorem?