The Riemann Zeta Function is most commonly defined as $$\zeta\left( s \right) = \sum\limits_{n = 1}^{\infty}\left[ \frac{1}{n^{s}} \right]$$ There is some sort of million dollar prize that involves proving the real part of complex number s must be $\frac{1}{2}$ for all nontrivial zeros. Of course this intregued me, because well, it's a million dollars. Odds are I won't solve it, but still. Anyway, I started looking at it and realized that you'd be raising a number to a complex power. This made no sense to me, so I went online and found Euler's formula that explains how that would work $$e^{i\pi}=-1$$ It turns out that the smallest nontrivial zeros is at about $\frac{1}{2}+14.1345i$, so I plugged it in to the zeta function. I used Desmos.com, and used separate summations for the real and imaginary parts. I expected to get zero. I did not get zero. In fact, the bigger I had the summation get, say, instead of summing to $1000000$, I'd sum to $1000000000$, the further off I would get from zero.
So tell me, how exactly are values for the Riemann Zeta Function computed?
One may note that when $\Re(s)\le1$,
$$\sum_{k=1}^\infty\frac1{k^s}\approx\int_1^\infty\frac1{x^s}\ dx\to\infty$$
Thus, we'll need a different representation of the zeta function. If we let $\eta(s)$ be the alternating form of the zeta function,
$$\eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s}$$
Then,
$$\zeta(s)-\eta(s)=\sum_{k=1}^\infty\frac{1+(-1)^k}{k^s}=\sum_{k=1}^\infty\frac2{(2k)^s}=2^{1-s}\zeta(s)$$
Thus, it follows that
$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$
By taking the Euler sum of the $\eta$ function, we get
$$\eta(s)=\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$
and thus, we reach a globally convergent form of the Riemann zeta function:
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$
and when testing for zeroes, the $\frac1{1-2^{1-s}}$ part is negligible.