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.



$$\sum_{n=1}^N\lambda(n)[N/n] =\sum_{n=1}^N\sum_{d|n}\lambda(d) $$

Next use this to conclude that the double sum equals the number of perfect squares not exceeding $N$

  • $\begingroup$ Can you explain how you convert it into a double sum? $\endgroup$ – Bosnia Mar 15 '15 at 13:26
  • $\begingroup$ 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 $[N/n]$ $\endgroup$ – ganeshie8 Mar 15 '15 at 13:36
  • $\begingroup$ More generally we have below for any number theoretic function $f$ : $$\sum_{n=1}^N f(n) [N/n] =\sum_{n=1}^N\sum_{d|n}f(d)$$ $\endgroup$ – ganeshie8 Mar 15 '15 at 13:44

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