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.

I would like to prove the following equality

$$\sum_{n=1}^N \mu^2(n) = \sum_{k=1}^{\sqrt{N}} \mu(k) \cdot \lfloor N / k^2 \rfloor$$

with N a square number.

Can anyone give me a hint?

p.s. I know already that

$$\frac{\zeta(s)}{\zeta(2s) } = \sum_{n=1}^{\infty}\frac{ \mu^2(n)}{n^{s}}$$

Perhaps this can help?

share|improve this question
2  
Notice that $\mu^2(n)$ is always $0$ or $1$ depending on wether $n$ is squarefree or not. So you can easily interpret the left-hand side... –  Joel Cohen May 4 '12 at 17:56
add comment

2 Answers

up vote 4 down vote accepted

Big hint: The left side counts something, and the right side counts the same thing via an inclusion-exclusion argument.

share|improve this answer
add comment

Hint: Since $$\sum_{d|n} \mu(d)=\left\{ \begin{array}{cc} 1 & \text{if }n=1\\ 0 & \text{otherwise} \end{array}\right\} $$ we have that

$$\sum_{d\leq N}\mu(d)^{2}=\sum_{d\leq N}\sum_{k^{2}|d}\mu(k).$$ Now, try switching the order of summation.

share|improve this answer
add comment

Your Answer

 
discard

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.