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The question I'm trying to solve is

$\sum_{n \le N} \phi(n) \lfloor \frac {N}{n} \rfloor $ = $\frac {N(N+1)}{2}$

for all natural numbers N where $\phi$ is Euler's totient function.

This is how far I've gotten:

$\sum_{n \le N} \phi(n) \lfloor \frac {N}{n} \rfloor $ = $\sum_{n \le N} \phi(n)$$\sum_{k \le \frac {N}{n}} 1$ = $\sum_{N \le k} \sum_{n|N} \phi (n)$ = $\sum_{N \le k} N$

Any suggestions or answers are very appreciated!

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You've made a good start writing

$$\biggl\lfloor \frac{N}{n}\biggr\rfloor = \sum_{k \leqslant N/n} 1,$$

but when you changed the order of summation you made a mistake. Just changing the order of summation would lead to

$$\sum_{n \leqslant N} \phi(n)\biggl\lfloor \frac{N}{n}\biggr\rfloor = \sum_{k \leqslant N} \sum_{n \leqslant N/k} \phi(n),$$

which doesn't make it obvious how to reach the goal either. The goal is reached by writing $m = n\cdot k$ and summing over $m$:

$$\sum_{n \leqslant N} \phi(n)\biggl\lfloor \frac{N}{n}\biggr\rfloor = \sum_{m \leqslant N} \sum_{n\cdot k = m} \phi(n)\cdot 1 = \sum_{m \leqslant N} \sum_{n \mid m} \phi(n) = \sum_{m \leqslant N} m = \frac{N(N+1)}{2}\,.$$

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  • $\begingroup$ thanks for the correction. I still don't know how to get it from the double sum to the final result, could you help me with the last step as well? $\endgroup$ – vfantina Mar 26 '18 at 15:07
  • $\begingroup$ You have seen $\sum_{n \mid m} \phi(n)$ before. You just need to remember what it was. It may help to think about what $h(m)$ must be for $\sum_{m \leqslant N} h(m) = \frac{1}{2}N(N+1)$. $\endgroup$ – Daniel Fischer Mar 26 '18 at 15:10
  • $\begingroup$ What I think I have at this point is $\sum_{m \le N} m$ so I assume that has to equal $\frac{N(N+1)}{2}$ $\endgroup$ – vfantina Mar 26 '18 at 15:20
  • $\begingroup$ Indeed. That's a rather famous formula. Is it more familiar when written $$\sum_{m = 1}^N m = \frac{N(N+1)}{2}\,?$$ $\endgroup$ – Daniel Fischer Mar 26 '18 at 15:24

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