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Let $f(n)$ be a multiplicative function defined by $f(p^a)=p^{a-1}(p+1)$, where $p$ is a prime number. How could I obtain a formula for $$\sum_{n\leq x} f(n)$$ with error term $O(x\log{x})$ and express the main term constant in terms of values of Riemann zeta function?

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What constant did you have in mind? –  Will Jagy Dec 10 '11 at 6:17
    
Should this have the homework tag? –  Greg Martin Dec 10 '11 at 7:21
    
This is actually not a homework problem. I am preparing my comprehensive exam for analytic number theory. This is one of the problems from past exams. –  Rob Dec 10 '11 at 17:50
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1 Answer

up vote 12 down vote accepted

The following is a more general presentation of a recurring idea which comes up when trying to find the mean value of certain multiplicative functions. Notice that the exact same steps would give us the result if $f(n)$ was replace by say $\phi(n)$, we would need only modify the calculation for $g(n)$ at the end.

Heuristics: Notice $f(n)\approx n$ so that $\frac{f(n)}{n}\approx 1$. For functions close to one, convolution with the Möbius function will be close to zero, so we can deal with it easily. Lets define $g(n)=(\mu*\frac{f(d)}{d})(n)=\sum_{d|n}\frac{f(d)}{d}\mu\left(\frac{n}{d}\right)$ so that $(1*g)(n)=\frac{f(n)}{n}$. The idea will be to rewrite everything in terms of $g$ since $g(n)$ will be small.

The main sum: We can write $$\sum_{n\leq x}f(n)=\sum_{n\leq x}n\frac{f(n)}{n}=\sum_{n\leq x}n\sum_{d|n}g(d)=\sum_{d\leq x}g(d)\sum_{n\leq x,\ d|n} n$$ $$=\sum_{d\leq x}dg(d)\sum_{n\leq \frac{x}{d}} n=\sum_{d\leq x}dg(d)\frac{\left[\frac{x}{d}\right]^2+\left[\frac{x}{d}\right]}{2}$$ $$=x^2\sum_{d\leq x}\frac{g(d)}{d}+O\left(x\sum_{d\leq x}|g(d)|\right).\ \ \ \ \ \ \ \ \ \ (1)$$ The error term comes from using $\left[\frac{x}{d}\right]=\frac{x}{d}+O(1)$, and then writing $$O\left(\sum_{d\leq x}d|g(d)|\left[\frac{x}{d}\right]\right)=+O\left(x\sum_{d\leq x}|g(d)|\right).$$

Calculating $g(d)$: Notice that $\frac{f(p^a)}{p^a}=\left(1+\frac{1}{p}\right)$. Then since $g(p^a)=\frac{f(p^a)}{p^a}-\frac{f(p^{a-1})}{p^{a-1}}$ for $a\geq 2$, we see that $g(p^a)=0$ when $a\geq 2$. When $a=1$ $g(p)=\frac{f(p)}{p}-1=\frac{1}{p}$. Hence $$g(n)=\frac{\mu(n)^2}{n}.$$

Putting this together: This means that $\sum_{d\leq x}|g(d)|=O(\log x)$, and that $\sum_{d\leq x}\frac{g(d)}{d}=\sum_{d\leq x}\frac{\mu^2(d)}{d^2}=\sum_{d=1}^\infty\frac{\mu^2(d)}{d^2}+O\left(\frac{1}{x}\right)$ so we have that by $(1)$ $$\sum_{n\leq x}f(n)=x^2 \sum_{d=1}^\infty \frac{\mu(d)^2}{d^2}+O(x\log x).$$

Using Euler products, $$\sum_{d=1}^\infty \frac{\mu(d)^2}{d^2}=\prod_p \left(1+\frac{1}{p^2}\right)=\prod_p \left(1-\frac{1}{p^4}\right)\prod_p \left(1-\frac{1}{p^2}\right)^{-1}=\frac{\zeta(2)}{\zeta(4)}.$$ Thus $$\sum_{n\leq x}f(n)=x^2 \frac{\zeta(2)}{\zeta(4)}+O(x\log x)=\frac{15}{\pi^2}x^2+O(x\log x).$$

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