Expanding Riemann Zeta Consider the Riemann Zeta Function
$\zeta(x) = 1 + 2^{-x} + 3^{-x} + 4^{-x}...$
Notice the following identity:
$a^{-x} = (e^{ln(a)})^{-x} = e^{-xln(a)}$
Therefore:
$\zeta(x) = 1 + 2^{-x} + 3^{-x} + 4^{-x}... = 1 + e^{-xln(2)} + e^{-xln(3)} + e^{-xln(4)}...$
Notice now that for each $e^{-xln(a)} $ we can expand it to find its mauclarin series:
$e^{-xln(a)} = 1 -xln(a) + \frac{x^2ln(a)^2}{2!} - \frac{x^3ln(a)^3}{3!}... $
Thus we can expand:
$\zeta(x) = 1 + e^{-xln(2)} + e^{-xln(3)} + e^{-xln(4)}...$
to
$\zeta(x) = 1 + $
$.......+ 1 -xln(2) + \frac{x^2ln(2)^2}{2!} - \frac{x^3ln(2)^3}{3!}... $
$.......+ 1 -xln(3) + \frac{x^2ln(3)^2}{2!} - \frac{x^3ln(3)^3}{3!}...$
$.......+ 1 -xln(4) + \frac{x^2ln(4)^2}{2!} - \frac{x^3ln(4)^3}{3!}...$
keeping this in mind:
suppose each of these terms was represented by the symbol H and we drew out a grid:
$H$  
$H$  $H$  $H$  $H$  $H$...
$H$  $H$  $H$  $H$  $H$...
$H$  $H$  $H$  $H$  $H$...
And we decided that we would approximate the riemann zeta function by picking up terms from this grid using some sort of space filling pattern
Would it be possible to construct a sequence of polynomials from the space filling algorithm whose roots steadily converged to real part 1/2 and maybe be able to make some limiting argument that as the the terms limit to infinity the real part limits to 1/2?
Just a thought... Would there be any flaws in this? Has this been done before?
 A: This may be of interest to you. The idea comes from Havil's book Gamma. First off, you don't want to use the zeta function for this, but the alternating zeta function. That's because the zeta function needs to be analytically continued to be valid for $0<\Re z<1$, and your current expansion will not be valid. The alternating zeta function is:
$$\zeta_a(z):=\sum_{n=1}^\infty \frac{(-1)^n}{n^z}=(1-2^{1-z})\zeta(z),$$
where the second equality is a fun exercise to prove (and has probably been answered dozens of times on this site). The point is that the function $\zeta_a(z)$ is now valid for  $0<\Re z<1$. Moreover, you can show that $\zeta(z)=0$ is equivalent to $\zeta_a(z)=0$. It's "obvious" from the current formula, but the point is to even say $\zeta(z)=0$ you need to first analytically continue it.
Now, write $z=\alpha+\beta i$ where $\alpha,\beta$ are real. As well, 
$$\frac{1}{n^z}=\frac{1}{n^\alpha}e^{-i\beta\ln n}=\frac{1}{n^\alpha}\left[\cos(\beta\ln n)-i\sin(\beta\ln n)\right]$$
Then $\zeta(z)=0$ is equivalent to saying $\zeta_a(z)=0$ which means
$$0=\sum_{n=1}^\infty\frac{(-1)^n}{n^\alpha}\left[\cos(\beta\ln n)-i\sin(\beta\ln n)\right],$$
where you now notice that splitting the two terms gives a purely real and purely complex sum. Thus $\zeta(z)=0$ is equivalent to simultaneously solving
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^\alpha}\cos(\beta\ln n)=0$$
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^\alpha}\sin(\beta\ln n)=0$$
for pairs $(\alpha,\beta)$. Then the Riemann Hypothesis is equivalent to $\alpha=1/2$ for all solutions [Edit, thanks to deoxygerbe: actually the correct statement is $\alpha=1/2$ or $\alpha=1$ iff the Riemann Hypothesis is true, this is because $z=1+2\pi i/\ln 2$ is a zero of $(1-2^{1-z})$, but by a result of de la Vallée-Poussin, $\zeta(s)$ has no zeros with real part equal to 1.] Notice that this simultaneous set of equations is a problem purely in the real numbers. You can truncate the series to your hearts content and try to look for numerical solutions, though I'm guessing (i'm not an expert) that this is a terrible way of approximating zeros. For that, look up Andrew Odlyzko and check out his numerical methods. 
Now, your idea applied to the above seems to be to try and expand $\cos(\beta\ln n)$. This is not wise. Notice that $\ln n$ is getting bigger and bigger as $n\rightarrow\infty$. So if you approximate $\cos(x)$ as a polynomial, Taylor or or otherwise, that approximation will start to fail miserably after $x$ gets large enough. 
