Prove that $G(n) = \sum_{d|n} \frac{n}{d}\tau(d)$ is multiplicative.

Question

I don't know how to prove this because it seems to me that $f(d)=\frac{n}{d}\tau(d)$ is not multiplicative.

Answer 1

The (Dirichlet) convolution of two multiplicative functions is multiplicative, i.e. $$ f(n)=\sum_{d|n}g(n/d)h(d) $$ is multiplicative if $g$, $h$ are. Both $g(n)=n$ and $\tau(n)=\sum_{d|n}1$ (if that's what you mean by $\tau$) are multiplicative, so their convolution is as well.