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I have just been told about this result, available as Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory (2nd edition)". It says:

Let $\alpha>0$. Suppose $a_n \ll n^{\alpha}$ and $A(x) \ll x ^{\delta} $ for some $\delta<1$, where $A(x) = \sum_{n\leq x} a_n$. Define $b_n = \sum_{d|n} a_d$. Then one has

$$\sum_{n\leq x} b_n = cx + O\Big(x^{(1-\delta)(1+\alpha)/(2-\delta)}\Big)$$ for some constant $c$.

Does anyone know who first came up with this result, or maybe it's just too straightforward to be attributed to anyone ?

The reason why i'm particularly interested in knowing the originator of this result is that, Murty's proof (which is on pages 262-263 of the aforementioned book) doesn't look quite right to me (but of course, i may be mistaken).

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  • $\begingroup$ What do you think is wrong with the proof presented? $\endgroup$ Jan 5, 2020 at 21:06
  • $\begingroup$ I quickly looked at the proof, and I could find no fault with it. I would submit that the result is straightforward enough so that particular attribution is necessary. $\endgroup$
    – 2734364041
    Jan 5, 2020 at 21:13
  • $\begingroup$ @Craig France, thanks for your comment. My concern is on the Dirichlet hyperbola formula that Murty's used. Is it the correct formula ? $\endgroup$
    – Q_p
    Jan 5, 2020 at 21:18
  • $\begingroup$ @2734364041, so may you please explain how the choice $y=x^{(1-\delta)/(2-\delta)}$ minimises the error term $x^{\delta}y^{1-\delta} + xy^{\delta-1}$ ? Also, plugging this choice of $y$ into the error term doesn't seem to yield $x^{(1-\delta-\delta^2)/(2-\delta)}$ ? $\endgroup$
    – Q_p
    Jan 5, 2020 at 21:26
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    $\begingroup$ the result in the book is incorrect as the computation of the powers is incorrect $\endgroup$
    – Conrad
    Jan 6, 2020 at 14:08

2 Answers 2

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It seems there is a typo in the application of the hyperbola method. Since $b_n=\sum_{d\mid n}a_d$, we have \begin{align} \sum_{n\le x}b_n &=\sum_{n\le x}\sum_{de=n}a_d=\sum_{de\le x}a_d\\ &=\sum_{\substack{de\le x\\ d\le y}}a_d+\sum_{\substack{de\le x\\ e\le x/y}}a_d-\sum_{de\le x\\ d\le y, e\le x/y}a_d. \end{align} If $A(x)=\sum_{n\le x}a_n$, then this can be written as \begin{align} \sum_{d\le y}a_d\left[\frac{x}{d}\right]+\sum_{e\le x/y}A\left(\frac{x}{e}\right)-A(y)\left[\frac{x}{y}\right]. \end{align} The author's assumption that $A(x)\ll x^{\delta}$ then gives the last two terms as $O(xy^{\delta-1})$, and the proof can be continued in the usual manner outlined in the text. Although, it appears that a sign was missed later on as well, which has some bearing on the final stated error (see Conrad's answer below).

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  • $\begingroup$ But isn't this a special case of Theorem 2.4.1 in the text? $\endgroup$
    – 2734364041
    Jan 6, 2020 at 3:57
  • $\begingroup$ Yes, you'll get the same thing if you set $f(n)=b_n$, $g(n)=a_n$, and $h(n)=1$ in Theorem 2.4.1. $\endgroup$ Jan 6, 2020 at 4:07
  • $\begingroup$ Thanks Craig ! So, the result is true but Murty's proof has a minor flaw, right ? $\endgroup$
    – Q_p
    Jan 6, 2020 at 7:43
  • $\begingroup$ Yeah, it happens. Was this result used in a paper? $\endgroup$ Jan 6, 2020 at 8:28
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    $\begingroup$ @as noted below, the result in the book is incorrect as the powers are not computed correctly $\endgroup$
    – Conrad
    Jan 6, 2020 at 14:11
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The computation of the powers is wrong and the result stated in the book is incorrect and it should be $cx+ O(x^{(1+\delta-\delta^2)/(2-\delta)})+O(x^{\frac{(1-\delta)(1+\alpha)}{2-\delta}})$

If $y=x^{(1-\delta)/(2-\delta)}$, $y^{\delta-1}=x^{-(1-\delta)^2/(2-\delta)}$, so $xy^{\delta-1}=x^{(1+\delta-\delta^2)/(2-\delta)}$ and it is not true that $\frac{1+\delta-\delta^2}{2-\delta} \le \frac{(1-\delta)(1+\alpha)}{2-\delta}$ in general, only for $\alpha \ge \delta + \frac{\delta}{1-\delta}$

Note that as stated, the result doesn't make sense because one can always increase $\delta$ in the hypothesis, while keeping $\alpha$ fixed, so in particular if $A(x) << x^{\delta}$ for a given $\delta < 1$, then $A(x) <<_{\epsilon} x^{1-\epsilon}$ for arbitrary $\epsilon >0$, hence we would get that $B(x) =cx + O_\epsilon(x^{\epsilon})$ under very general conditions and I am sure lots of counterxamples to that can be found

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  • $\begingroup$ You're right, it looks like a sign was missed later on. Also, the $O(y^{\alpha+1})$-term looks like it could have been replaced by $O(y^{\delta})$ since $\delta<1<1+\alpha$ ... that would make the error term $O(xy^{\delta-1})$. $\endgroup$ Jan 6, 2020 at 16:19

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