Let $f$ be an arithmetic function. If $$\lim_{x \rightarrow \infty}\frac{\sum_{n \leq x} f(n)}{x} = M \neq 0, $$ deduce the asymtotic formula for $$\sum_{n \leq x} f(n)n^{ia}$$ where $a$ is a real number constant.

I am not sure about what the question is asking. We have notation of asymptotic $f \sim g := \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1.$

So I have to find $g$ such that $$\sum_{n \leq x} f(n)n^{ia} \sim g.$$

Is it equivalent to $$\sum_{n \leq x} f(n)n^{ia} = h_1(x) + O(h_2(x)) ?$$ I usually see many theorem in analytic number theory write the terms in the way $$\sum_{n \leq x} f(x) = mainterm + O(something).$$ Is this thing also included as asymptotic formular ?

For the prove, I plan to use Abel summation formula (https://en.wikipedia.org/wiki/Abel%27s_summation_formula) to write

$$\sum_{n \leq x} f(n)n^{ia} = h(x) + O(something)$$ by setting $$a_n = f(n), A(t) = \sum_{n\leq t}a_n, \phi(t) = t^{ia}.$$

But I am not sure if Abel formular valid for complex function, and more over what is the (complex) derivative of $t^{ia} ?$

  • $\begingroup$ Abel formula is valid for complex-valued functions as well. And of course the derivative of $\phi : (0, \infty) \to \Bbb{C}$ given by $\phi(t) = t^{ia}$ is $ia t^{ia-1}$ as you expect. Finally, if I computed correctly, Abel formula leads to $$\sum_{n \leq x} f(n) n^{ia} \quad \sim \quad \frac{M}{ia+1} x^{ia+1} $$ as $x \to \infty$. $\endgroup$ Oct 3, 2016 at 13:06
  • $\begingroup$ @SangchulLee Thank you for your response. Could you please write what you did ? Just some steps is fine. I will do Abel summation, but it is good if I have something to check when I do it. Actually, I only use Abel and get to $mainterm + O(something)$, not sure how to obtain $\sim$ there. $\endgroup$
    – Both Htob
    Oct 3, 2016 at 13:12

1 Answer 1


Abel's formula gives you

$$ \sum_{n \leq x} f(n) n^{ia} = A(x)x^{ia} - ia \int_{1}^{x} A(u) u^{ia-1} \, du. \tag{*}$$

Now using the assumption, let us write $A(u) = (M + \epsilon(u))u$ where $\epsilon(u) \to 0$ as $u \to \infty$. Then

\begin{align*} \int_{1}^{x} A(u) u^{ia-1} \, du &= \int_{1}^{x} (M + \epsilon(u)) u^{ia} \, du \\ &= \frac{M}{ia+1}(x^{ia+1} - 1) + \int_{1}^{x} \epsilon(u) u^{ia} \, du \end{align*}

We claim that the last integral is $o(x)$ as $x \to \infty$. Indeed, one can adopt the idea of the Cesaro-mean to check that

$$ \frac{1}{x} \left|\int_{1}^{x} \epsilon(u) u^{ia} \, du\right| \leq \frac{1}{x} \int_{1}^{x} |\epsilon(u)| \, du \to 0 \quad \text{as} \quad x \to \infty. $$

Thus we have

$$\int_{1}^{x} A(u) u^{ia-1} \, du = \frac{M}{ia+1}x^{ia+1} + o(x). $$

Plugging this back to $\text{(1)}$ and using the relation $A(x) = Mx + o(x)$ gives

$$ \sum_{n \leq x} f(n) n^{ia} = \frac{M}{ia+1}x^{ia+1} + o(x), $$

which is essentially equivalent to

$$\sum_{n \leq x} f(n) n^{ia} \sim \frac{M}{ia+1}x^{ia+1}. $$

  • $\begingroup$ Thank you very much !! So $\sim$ is connected to little o ? I do not know that. Is it also connect to big O ? Might be not since the definition of Big O and little o is considerably different. $\endgroup$
    – Both Htob
    Oct 3, 2016 at 13:35
  • $\begingroup$ @BothHtob, The connection comes from the fact that $f(x) \sim g(x)$ can be written as $f(x)/g(x) = 1 + o(1)$. So you can freely move from $f(x) \sim g(x)$ to $f(x) = g(x) + o(g(x))$. The other direction is also guaranteed given that $g(x)$ never vanishes as $x \to \infty$. In our case, since $M/(ia+1)$ is non-zero and $|x^{ia+1}| = x$, we have $o(\frac{M}{ia+1}x^{ia+1}) = o(x)$. $\endgroup$ Oct 3, 2016 at 14:11

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