Show that $\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^s}=\frac{\z eta(s-1)}{\zeta(s)}.$
$\zeta(s)\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^s}= \sum_{m=1}^{\infty} \frac{1}{m^s}\sum_{n=1}^{\infty}\frac{\varphi(n)}{ n^s}=\sum_{k=1}^{\infty}\frac{1}{k^s}\sum_{d \mid k} \varphi(d)=\sum_{k=1}^{\infty} \frac{1}{k^{s-1}}=\zeta(s-1). \ \ \ \Box$