Let $s=\sigma+i t$, where $\sigma$ and $t$ are real, and for $\sigma>1$ let

$$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}}$$

Prove that $\zeta(s)$ has no zeros in the half plane $\sigma>1$. Show also that for $\sigma>1$,

$$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}$$

where $\mu$ denotes the Möbius function. Assuming that $\zeta(s)-\frac{1}{s-1}$ has an analytic continuation to the half plane $\sigma>0$, show that if $s=1+i t$, with $t \neq 0$, and $\zeta(s)=0$ then $s$ is at most a simple zero of $\zeta$.