The Riemann Zeta function represented by the following series, valid for $\Re(s)>0$,
$$ \zeta(s)=\frac{1}{1-2^{1-s}}\cdot\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^s} $$
appears to have zeros along the line $\Re(s)=1/2$.
I know that proving the zeros all lie on that line is an open problem, however ...
Question: Is it relatively simple to prove there are zeros in the domain $0<\Re(s)<1$?
The following approach requires awareness of the functional equation and $\xi(s)$. Using the fact that $\xi(s)$ is an entire function of order one, one can deduce that if $\zeta(s)$ only has finitely many (or no) zeros in the critical strip then $|\log\xi(s)|\ll|s|$ as $|s|\to\infty$. However, using Stirling's approximation for Gamma function, one can see that $|\log\xi(s)|\gg|s|\log|s|$ when $|s|\to\infty$ in the positive real direction, reaching a contradiction.
The strength of the above approach is that not only does it show that $\zeta(s)$ has nontrivial zeros in $0<\Re(s)<1$ but also deduces that there are infinitely many of them.
For a more detailed account, you might want to have a look at Ram Murty's Problems in Analytic Number Theory. Hope this can address your concern!