Let $f$ be an arithmetic function such that $\sum_{n \leq N} f(n) n^{-\frac{1}{2}} \sim N$ as $N \rightarrow \infty$. Prove that
\begin{equation} \sum_{n \leq N} f(n) \sim \frac{2}{3} N^{\frac{3}{2}} \quad \text{ as } N \rightarrow \infty. \end{equation}
I have an issue with understanding how to solve such types of problem (perhaps someone can point me to some reference). I was thinking of applying Euler's Summation Formula, Summation by Parts (but they require that $f$ is $C^{1}$), and the Convolution Method. But I can't seem to proceed. Also how are such expressions of the form $\sum_{n \leq N} f(n) n^{-\frac{1}{2}} \sim N$ called? Thanks in advance!
Starting with a sum of terms $f(n)/\sqrt{n}$ is weird. Call that $a_n$, so your task can be presented in a more elegant form: if $\sum_{n \leq N} a_n \sim N$ as $N \to \infty$ then $\sum_{n \leq N} a_n\sqrt{n} \sim (2/3)N^{3/2}$ as $N \to \infty$. (Asymptotic relations below are understood to be taking place "as $N \to \infty$" so I won't repeat that each time.)
Intuitively, if $\sum_{n \leq N} a_n \sim N$ then we imagine this is because $a_n \sim 1$ (that would explain the hypothesis in a simple way), and with $a_n = 1$ we'd get $\sum_{n \leq N} a_n\sqrt{n} = \sum_{n \leq N} \sqrt{n} \sim (2/3)N^{3/2}$, so the conclusion looks plausible. Do you agree that $\sum_{n \leq N} n^\alpha \sim N^{\alpha+1}/(\alpha+1)$ when $\alpha > -1$, or at least when $\alpha > 0$?
Aim for a broader result than your task: if $\sum_{n \leq N} a_n \sim N$ then $\sum_{n \leq N} a_nn^\alpha \sim N^{\alpha+1}/(\alpha+1)$ when $\alpha > 0$. Your task is the case $\alpha = 1/2$. All you need for this is summation by parts (why do you say summation by parts requires a function is $C^1$?) plus the following property of asymptotic partial sums: if $b_n$ and $c_n$ are positive sequences with $b_n \sim c_n$ as $n \to \infty$ and $\sum_{n \leq N} b_n$ diverges (or $\sum_{n \leq N} c_n$ diverges) then $\sum_{n \leq N} b_n \sim \sum_{n \leq N} c_n$ as $N \to \infty$.