On $\sum (1/\sqrt{n})\cos (t_0 \log n)$ and $\sum (1/\sqrt{n})\sin (t_0 \log n)$, from a zero of $\zeta (s)$ of the form $s_0=(1/2)+it_0$ I assume (as hypothesis, for questions too) that $s_0$ a fixed (nontrivial) zero of the Riemann zeta function $\zeta(s)=\sum_{n=1}^\infty 1/n^s$ has the form $(1/2)+it_0$, Thus for a positive real  $t_0$, since $n^{s_0}=\sqrt{n}e^{it_{0}\log n}$, multiplying denominator and numerator by $e^{-it_{0}\log n}$ in $0=\sum_{n=1}^{\infty} 1/n^{s_0}$, 

Question 1. can I write from this zero 
  $$\sum_{n=1}^{\infty} (1/\sqrt{n})\cos (t_0 \log n)=i\cdot\sum_{n=1}^{\infty} (1/\sqrt{n})\sin (t_0 \log n)?$$

And from previous identity, taking alternatively, real and imaginary parts, 

Question 2. can I write 
  $$\sum_{n=1}^{\infty} (1/\sqrt{n})\cos (t_0 \log n)=\sum_{n=1}^{\infty} (1/\sqrt{n})\sin (t_0 \log n)=0?$$

It is, what type of convergence can I assume and claim in previous computations? I want to understand thess computations to understand what type of convergence  there is around a zero of $\zeta (s)$. Thanks in advance.
References:
[1] Wikipedia, Riemann Zeta Function.
 A: No, you can't. The definition of zeta function $$\zeta\left(s\right)=\sum_{n\geq1}\frac{1}{n^{s}}
 $$ holds only for $\textrm{Re}\left(s\right)>1
 $. There are other definition of zeta as a series in bigger areas of the complex plain, for example $$\zeta\left(s\right)=\frac{1}{1-2^{1-s}}\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n^{s}}
 $$ holds for $\textrm{Re}\left(s\right)>0
 $ or $$\zeta\left(s\right)=\frac{1}{s-1}\sum_{n\geq0}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k}}{\left(1+k\right)^{s}}
 $$ holds on the entire complex plane except at $s=1
 $. Anyway, if we take $n^{s_{0}}=\sqrt{n}e^{it_{0}\log\left(n\right)}
 $ we have $$\frac{1}{n^{s_{0}}}=\frac{1}{\sqrt{n}e^{it_{0}\log\left(n\right)}}=\frac{1}{\sqrt{n}\left(\cos\left(t_{0}\log\left(n\right)\right)+i\sin\left(t_{0}\log\left(n\right)\right)\right)}
 $$ and in general it is not equal to $$\frac{1}{\sqrt{n}}\left(\cos\left(t_{0}\log\left(n\right)\right)+i\sin\left(t_{0}\log\left(n\right)\right)\right)
 $$ as you write.
