TITLE: On Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory."
QUESTION [5 upvotes]: 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).

REPLY [3 votes]: 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