Proof on Euler totient function

Solution 1:

When $n $ is even, then we will have terms $(d,2d, 4d,\cdots, 2^kd)$, such that $d$ is odd and $d, 2d, 4d,\cdots, 2^kd |n$. Also $\frac{n}{2^kd}$ is odd, but for other terms it is even. Now, $$ (-1)^{n/d}\phi(d)+(-1)^{n/2d}\phi(2d)+\cdots+(-1)^{n/2^{k-1}d}\phi(2^{k-1}d)+(-1)^{n/2^kd}\phi(2^kd) =\phi(d)+\phi(d)+2\phi(d)+\cdots+2^{k-2}\phi(d)-2^{k-1}\phi(d) =\phi(d)+(2^{k-1}-1)\phi(d)-2^{k-1}\phi(d)=0 $$ Hence for even $n$, $\sum_{d|n}(-1)^{n/d}\phi(d)=0.$

When $n $ is odd, $(-1)^{n/d}=-1$. Hence, for odd $n$, $\sum_{d|n}(-1)^{n/d}\phi(d)=-\sum_{d|n}\phi(d)=-n$.

Solution 2:

You may be interested to know that since $\sum_{d|n}\varphi(d)=n$ we have

$$\zeta(s) \sum_{n\ge 1}\frac{\varphi(n)}{n^s} = \zeta(s-1)$$

or $$L_1(s)= \sum_{n\ge 1}\frac{\varphi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}.$$

Furthermore $$\left(1-\frac{2}{2^s}\right)\zeta(s) = \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^s}$$ so that $$L_2(s) = \sum_{n\ge 1}\frac{(-1)^n}{n^s} = -\left(1-\frac{2}{2^s}\right)\zeta(s) .$$

We thus have $$L_1(s) L_2(s) = \sum_{n\ge 1}\frac{1}{n^s}\sum_{d|n} (-1)^{n/d}\varphi(d) = -\left(1-\frac{2}{2^s}\right)\zeta(s-1) .$$

Now when $n$ is odd then $1/n^s$ appears just once in the bracketed term and we obtain $-n.$ When $n$ is even it appears in both terms with a contributions of $-n$ from the first and $2\times n/2$ from the second, these sum to zero. This is the claim. (The Dirichlet series converge for $\Re(s) \ge 2.$)