nt.number theory - Inequality with Euler's totient

As a matter of fact your sum diverges, a little manipulation shows that

$$sum_{n leq X} frac{(-1)^n}{phi(n)} = sum_{n leq X, 2|n} frac{1}{phi(n)} - sum_{n leq X, (n,2)=1} frac{1}{phi(n)} = sum_{n leq X/2} frac{1}{phi(2n)} -sum_{n leq X,(n,2)=1}frac{1}{phi(n)}$$
The above equala to
$$sum_{n leq X/2, (n,2)=1} frac{1}{phi(2n)} + sum_{n leq X/2, 2|n} frac{1}{phi(2n)} - sum_{n leq X, (n,2)=1} frac{1}{phi(n)}$$ By multiplicativity of $phi(n)$ we have $phi(2n) = phi(n)$ when $(n,2)=1$. Thus the first sum above is a sum over $1/phi(n)$ and the above equation simplifies to
$$- sum_{X/2 < n leq X, (n,2)=1} frac{1}{phi(n)} + sum_{n leq X/4} frac{1}{phi(4n)}$$ It follows that
$$sum_{n leq X} frac{(-1)^n}{phi(n)}= -sum_{X/2 < n leq X, (n,2)=1} frac{1}{phi(n)} + sum_{n leq X/4} frac{1}{phi(4n)} sim c cdot log{X}$$
because the first sum on the right converges to a constant, while the second sum on the right is asymptotically $c cdot log{X}$.

EDIT: Put details, erased mention of an earlier confusion about $(-1)^{n+1}$ not being a multiplicative function :P (it is!)

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