Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

up vote 13 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 $$\sum_{n\leq x}f(n)=x^2 \frac{\zeta(2)}{\zeta(4)}+O(x\log x)=\frac{15}{\pi^2}x^2+O(x\log x).$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.