I have to prove $$\sum_{n=1}^N\lambda(n)[N/n]=[\sqrt{N}]$$ I tried using the approach in this question but I don't know how I'll get $\sqrt{N}$. Please help.
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Similar: http://math.stackexchange.com/questions/8002, http://math.stackexchange.com/questions/332588 – Bart Michels Mar 15 '15 at 14:33
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HINT
$$\sum_{n=1}^N\lambda(n)[N/n] =\sum_{n=1}^N\sum_{dn}\lambda(d) $$
Next use this to conclude that the double sum equals the number of perfect squares not exceeding $N$


Notice that for a fixed $n$, on the left hand side the function $\lambda(n)$ is getting evaluated $[N/n]$ times. On the right hand side also the specific term $\lambda(n)$ appears exactly $[N/n]$ times in the expanded sum because the number of integers divisible by $n$ is $[N/n]$ – AgentS Mar 15 '15 at 13:36

More generally we have below for any number theoretic function $f$ : $$\sum_{n=1}^N f(n) [N/n] =\sum_{n=1}^N\sum_{dn}f(d)$$ – AgentS Mar 15 '15 at 13:44