Analytic continuation for $\zeta(s)$ using finite sums? $\zeta(s)$ converges for $\sigma >1$ but not for $\sigma =1/2.$
But for some reason for $s = 1/2 + i t $ and fixed finite $N,~$ $\zeta_N(s) =\sum_{n=1}^N\frac{1}{n^s}$ is very close to $\zeta(s)$ as found using analytic continuation for $t$ in some range
$$f_1(N) < t < f_2(N). $$
For example, for $N= 1000$ a plot of $\zeta_N(s)$ gives a zero at approximately $t= 568.9,$ and the 319th zeta zero is about $t=568.924$ (per Mathematica). For $N=10000,$ graphically there is a zero at approximately $1783.07,$ and this agrees with Mathematica's 1321st zero, about $1783.08.$ 
Mathematica has a built-in analytic continuation for $\zeta(s),$ but I don't think that is being engaged here. 

Does this property evaporate for very large $N$ or is there a reason we should in principle be able to approximate $\zeta(s)$ at $\sigma = 1/2$ for finite $N$ on a suitable range of $t?$  

Since this property may not hold for large $N$ there is little point in broadening the question. 
 A: Let $f\left(x\right)\in C^{1}\left[a,b\right]$, then holds $$\sum_{a<n\leq b}f\left(n\right)=\int_{a}^{b}f\left(x\right)dx+\int_{a}^{b}\left(x-\left[x\right]-\frac{1}{2}\right)f'\left(x\right)dx+\left(a-\left[a\right]-\frac{1}{2}\right)f\left(a\right)-\left(b-\left[b\right]-\frac{1}{2}\right)f\left(b\right)$$ where $\left[x\right]$ is the floor function. Then if we take $s=\sigma+it$, $\sigma>1$ and with a suitable choice of $a,b$ we have $$\zeta\left(s\right)-\sum_{n=1}^{N}\frac{1}{n^{s}}=s\int_{N}^{\infty}\frac{\left[x\right]-x+\frac{1}{2}}{x^{s+1}}dx+\frac{N^{s-1}}{s-1}-\frac{1}{2N^{s}}$$ and the result holding for analytic continuation for $\sigma>0$. Hence we have 
\begin{align*}
\zeta\left(s\right)-\sum_{n=1}^{N}\frac{1}{n^{s}}= & O\left(t\int_{N}^{\infty}\frac{1}{x^{\sigma+1}}dx\right)+O\left(\frac{N^{1-\sigma}}{t}\right)+O\left(N^{-\sigma}\right)\\
= & O\left(\frac{t}{\sigma N^{\sigma}}\right)+O\left(\frac{N^{1-\sigma}}{t}\right)+O\left(N^{-\sigma}\right)
\end{align*}
 and if we take $N=\left[t\right]$ we have $$\zeta\left(s\right)-\sum_{n=1}^{N}\frac{1}{n^{s}}=O\left(1\right).$$ 
