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

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.

• What does $\sum f(x)$ even mean? Commented Jul 14, 2019 at 1:22
• Under any reasonable interpretation of this question, it seems to me it should be settled by the limit comparison test. Commented Jul 14, 2019 at 1:23

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.

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?