Define $f \sim g$ to mean $\lim_{x \to \infty}\frac{f(x)}{g(x)}=1$.
I am trying to either prove or disprove that $f \sim g$ $\iff$ $\sum_{k=1}^{x}f(k) \sim \sum_{k=1}^{x} g(k)$ provided $f$ and $g$ are positive and unbounded for large $x$.
I showed that $\leftarrow$ does not hold, but dropped the assumption that $f$ and $g$ remain unbounded (I took $f(x)=\frac{1}{x(x+1)}$ and $g(x)=\frac{1}{x(x+1)(x+2)}$, in which case the partial sums are $\sim$, but $f$ and $g$ aren't.) For $\rightarrow$, I tried to use Abel's summation formula and attempted to show that $\sum_{k=1}^{x}f(k) -\sum_{k=1}^{x} g(k) \rightarrow 0$, but got nowhere.
Under the given assumptions, the implication $$f(x) \sim g(x) \implies \sum_{k = 1}^x f(k) \sim \sum_{k = 1}^x g(k)$$ holds, but the implication in the other direction does not hold.
To see the latter, consider $f(x) = (2 + \cos (\pi x))\cdot x$ and $g(x) = (2 - \cos (\pi x))\cdot x$. Then $\frac{f(k)}{g(k)}$ alternates between $3$ and $1/3$, so $f(x) \nsim g(x)$, but $$\sum_{k = 1}^x f(k) = x^2 + O(x) \sim \sum_{k = 1}^x g(k) = x^2 + O(x).$$
To see that the implication above holds, consider an arbitrary $\varepsilon > 0$. By the assumption $f(x) \sim g(x)$ there is an $x_{\varepsilon} \in \mathbb{N}$ such that $$(1-\varepsilon) f(x) \leqslant g(x) \leqslant (1 + \varepsilon)f(x)$$ for all $x \geqslant x_{\varepsilon}$. Then for $x > x_{\varepsilon}$ we have $$(1 - \varepsilon)\sum_{k = 1}^x f(k) - (1 - \varepsilon)\sum_{k = 1}^{x_{\varepsilon}} f(k) + \sum_{k = 1}^{x_{\varepsilon}} g(k) \leqslant \sum_{k = 1}^x g(k) \leqslant (1 + \varepsilon)\sum_{k = 1}^x f(k) - (1 + \varepsilon) \sum_{k = 1}^{x_{\varepsilon}} f(k) + \sum_{k = 1}^{x_{\varepsilon}} g(k).$$ Since the sums up to $x_{\varepsilon}$ are independent of $x$, and $\sum_{k = 1}^x f(k) \to \infty$ (that is what we need, $f$ need not be unbounded itself) it follows that $$1 - 2\varepsilon < \frac{\sum_{k = 1}^x f(k)}{\sum_{k = 1}^x g(k)} < 1 + 2\varepsilon$$ for sufficiently large $x$.