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I want to prove that $$ \sum_{k=1}^n \left(\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n-1}{k}\right\rfloor\right)\varphi(k)=n. $$

In a book I found this exercise:

Prove that $$ \sum_{k\leq n} \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) =\frac{n(n+1)}{2}. $$

Which is proven here. I proved it this way. Let $$ \sum_{k\leq n} \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) =: u_n. $$ then $$ u_n-u_{n-1} = \sum_{k=1}^n \left(\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n-1}{k}\right\rfloor\right)\varphi(k)=n. $$ and hence $$ u_n-u_1=\sum_{j=1}^{n-1} (u_{j+1}-u_{j})=\sum_{j=1}^{n-1} (j+1)=\frac{n(n+1)}{2}-1. $$ As $u_1=\varphi(1)=1$, $$ u_n=\frac{n(n+1)}{2}. $$

All that remains to show is the equality above, I am very sure that it is true, however I need some help. Any hints are welcome.

Bill Dubuque
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calculatormathematical
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    Hint:Recall that $\sum_{k\mid n} \varphi(k) = n$, which looks similar (same form) as yours. $\quad$ Is it the same? Your coefficient is either 0 or 1. When is it 1? – Calvin Lin Sep 09 '24 at 16:03

1 Answers1

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Hint: Recall that $ \sum_{k\mid n } \varphi(k) = n$.

Show that this is identical to your equation, namely

$$\left\lfloor \frac{n}{k}\right\rfloor - \left\lfloor \frac{n-1}{k}\right\rfloor = \begin{cases} 1 & k \mid n \\ 0 & k \not\mid n \\ \end{cases}$$

Calvin Lin
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