zeta function, a nontrivial zero that is not on critical line

Question

The anylitical continued version of the zeta function is the eta function divided by $(1-2/2^s)$, since $(1-2/2^s)$ is never $\infty$, The zeros of $\zeta(s)$ the must be the same as the zeroes of $\eta(s)$. I came up with $$\eta(x+iy)=i\sum_1^{\infty}n^{-x}{(-1)}^{n+1}\sin(-y\ln(n))+\sum_1^{\infty}n^{-x}{(-1)}^{n+1}\cos(y\ln(n))$$ I plugged in $1+\approx18.125i$ and got $0$. But i plugged that to wolfram alpha and got a different answer (zeta). But the It is a zero of the eta function. Where is my logic flawed.

Answer 1

Since

$$ \eta(s) = \left(1-2^{1-s}\right)\zeta(s), $$

he $\eta$ function has all of the zeros of $\zeta$ as well as all of the zeros of $1-2^{1-s}$. That zero you found is actually

$$ 1+\frac{4\pi i}{\ln 2} \approx 1 + 18.1294i, $$

which is, indeed, a zero of $1-2^{1-s}$.