The totient sum function has an identity:

$\displaystyle\sum_{k=1}^{N}\varphi(k) = \sum_{k=1}^N {\rm M}(\lfloor\frac{N}{k}\rfloor)k $

$\varphi(k)$ is the Euler totient function, and $M$ is the Mertens function $\displaystyle M(N)=\sum_{k=1}^N \mu (k)$ where $\mu$ is the Moebius function.

My question: What is the Mertens function summation equivalent of $\displaystyle\sum_{k=1}^{N}k\varphi(k) $ and $\displaystyle\sum_{k=1}^{N}k^2\varphi(k)$?

  • 2
    $\begingroup$ See math.stackexchange.com/questions/316376/… and math.stackexchange.com/questions/317482/… the latter deleted a few minutes ago. $\endgroup$
    – Will Jagy
    Mar 1, 2013 at 4:57
  • $\begingroup$ @WillJagy Yes, that was my question. It had not been getting much response. $\endgroup$ Mar 1, 2013 at 5:05
  • 1
    $\begingroup$ Why do you think there is anything to be said about this? And what are you doing? $\endgroup$
    – Will Jagy
    Mar 1, 2013 at 5:19
  • $\begingroup$ Why why why why do you refuse, despite repeated urging, to link your questions to your earlier, related questions? Why does someone else always have to do this for you? $\endgroup$ Mar 1, 2013 at 11:43
  • $\begingroup$ @GerryMyerson I spoke with joriki about it; won't happen again. Sorry for not linking the questions. $\endgroup$ Mar 1, 2013 at 15:09

1 Answer 1


Since $\varphi(n)=n\sum_{d|n}\frac{\mu(d)}{d},$ by switching the order of summation we have that for fixed $l$ $$\sum_{n\leq x}\varphi(n)n^{l}=\sum_{kd\leq x}\mu(d)k^{l}d^{l}k$$

$$=\sum_{k\leq x}k^{l+1}\sum_{d\leq\frac{x}{d}}\mu(d)d^{l}.$$ Now, $$\sum_{d\leq y}\mu(d)d^{l}=\int_{0}^{y}t^{l}d\left(M(t)\right)=M(y)y^{l}-l\int_{0}^{y}M(t)t^{l-1}dt,$$ so you can write $$\sum_{n\leq x}\varphi(n)n^{l}=\sum_{k\leq x}k^{l+1}\left(\frac{x}{k}\right)^{l}M\left(\frac{x}{k}\right)-l\int_{0}^{\frac{x}{k}}t^{l-1}M(t)dt.$$ This may be rearranged as

$$x^{l}\sum_{k\leq x}kM\left(\frac{x}{k}\right)-l\sum_{k\leq x}k^{l+1}\int_{0}^{\frac{x}{k}}t^{l-1}M(t)dt,$$ which is equal to

$$x^{l}\sum_{k\leq x}kM\left(\frac{x}{k}\right)-l\int_{0}^{x}t^{l-1}\sum_{k\leq x}k^{2}M\left(\frac{t}{k}\right)dt.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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