In Apostol’s book, Theorem 3.6 (p.61) states a result concerning the average order of ${\sigma _\alpha }(n)$, where $\alpha < 0$
I am including an outline of Apostol’s approach, I hope I have typeset it correctly!
In Apostol’s approach, the positive parameter $\beta$ is defined, where $\beta = - \alpha$
The result being proved is
$$\sum\limits_{n\, \le x}^{} {{\sigma _{ - \beta }}} (n) = \,\,\varsigma (\beta + 1)x + O({x^\delta }),\,\,\,\,\beta \ne 1$$ $$\sum\limits_{n\, \le x}^{} {{\sigma _{ - \beta }}} (n) = \,\,\varsigma (2)x + O(\log x),\,\,\,\,\beta = 1$$ where $\delta = \max \{ 0,1 - \beta \} $.
In his proof, Apostol applies the following asymptotic formulas: $$\sum\limits_{n \le x} {\frac{1}{n}} \,\, = \,\,\log x\,\, + \,\,\gamma \,\, + \,\,O\left( {\frac{1}{x}} \right);$$ $$\sum\limits_{n \le x} {\frac{1}{{{n^s}}}} \, = \frac{{{x^{1 - s}}}}{{1 - s}} + \varsigma (s) + O({x^{ - s}}),\,\,s > 0,\,\,s \ne 1\,;$$ $$\sum\limits_{n \le x} {{n^\alpha }} \, = \frac{{{x^{\alpha + 1}}}}{{\alpha + 1}} + O({x^\alpha }),\,\,\alpha \ge 0.$$
His approach is then as follows:
$$\sum\limits_{n \le x} {{\sigma _{ - \beta }}(n) = \sum\limits_{n \le x} {\sum\limits_{d|n} {\frac{1}{{{d^\beta }}}} } } = \sum\limits_{n \le x} {\frac{1}{{{d^\beta }}}} \sum\limits_{q \le x/d} 1 $$ $$ = \sum\limits_{d \le x} {\frac{1}{{{d^\beta }}}\left\{ {\frac{x}{d} + O(1)} \right\}} = x\sum\limits_{d \le x} {\frac{1}{{{d^{\beta + 1}}}}} + O\left( {\sum\limits_{d \le x} {\frac{1}{{{d^\beta }}}} } \right)$$ Apostol states that the last term is $O(\log x)$ if $\beta = 1$ and $O\left( {{x^\delta }} \right)$ if $\beta \ne 1$.
Once again, using asymptotic formula, Apostol concludes his proof by stating $$x\sum\limits_{d \le x} {\frac{1}{{{d^{\beta + 1}}}}} = \frac{{{x^{1 - \beta }}}}{{1 - \beta }} + \varsigma (\beta + 1)x + O\left( {{x^{ - \beta }}} \right) = \varsigma (\beta + 1)x + O\left( {{x^{1 - \beta }}} \right)$$
When working through this theorem, I attempted to prove the result myself using the given asymptotic formula. My working did not get to the result; I am just wondering where I may have gone wrong. It may indeed be one of those “tricks” to formulate $\sum\limits_{n \le x} {{\sigma _{ - \beta }}(n)} $ in the “right” way so that key terms are not lost in big-oh representations, etc.
This is how I approached it:
$$\sum\limits_{n \le x} {{\sigma _{ - \beta }}(n)} \,\, = \,\,\sum\limits_{d \le x} {\sum\limits_{q \le {\textstyle{x \over d}}} {{q^{- \beta }}} } = \,\,\sum\limits_{d \le x} {\sum\limits_{q \le {\textstyle{x \over d}}} {\frac{1}{{{q^\beta }}}} } $$ $$\sum\limits_{d \le x} {\left\{ {\frac{{{{(x/d)}^{1 - \beta }}}}{{1 - \beta }} + \varsigma (\beta ) + O\left( {{{\left( {{\textstyle{x \over d}}} \right)}^{ - \beta }}} \right)} \right\}} $$ $$ = \frac{{{x^{1 - \beta }}}}{{1 - \beta }}\sum\limits_{d \le x} {\frac{1}{{{d^{1 - \beta }}}}} + [x]\varsigma (\beta ) + \sum\limits_{d \le x} {O\left( {{{\left( {{\textstyle{d \over x}}} \right)}^\beta }} \right)} $$ $$ = \frac{{{x^{1 - \beta }}}}{{1 - \beta }}\sum\limits_{d \le x} {\frac{1}{{{d^{1 - \beta }}}}} + [x]\varsigma (\beta ) + O\left( {\frac{1}{{{x^\beta }}}\sum\limits_{d \le x} {{d^\beta }} } \right)$$ $$ = \frac{{{x^{1 - \beta }}}}{{1 - \beta }}\sum\limits_{d \le x} {\frac{1}{{{d^{1 - \beta }}}}} + [x]\varsigma (\beta ) + O\left( {\frac{1}{{{x^\beta }}}\left\{ {\frac{{{x^{\beta + 1}}}}{{\beta + 1}} + O({x^\beta })} \right\}} \right)$$
The "${\frac{1}{{{d^{1 - \beta }}}}}$" on the left is problematic, because the asymptotic formula used depends on if $\beta > 1$ or $\beta < 1$.
At this point I ran out of steam and looked at Apostol’s approach.
But I am still intrigued… is my analysis incorrect, or is it just an alternative approach that ends up with a different, but ultimately unhelpful, result?
And are there any clues that would lead to breaking down of the initial expression like this:
$$\sum\limits_{n \le x} {{\sigma _{ - \beta }}(n) = \sum\limits_{n \le x} {\sum\limits_{d|n} {\frac{1}{{{d^\beta }}}} } } = \sum\limits_{n \le x} {\frac{1}{{{d^\beta }}}} \sum\limits_{q \le x/d} 1 $$
My first post, so I hope I have made myself clear and not violated any rules or contravened accepted formats and styles!
Thanks!