sum of
1/2+1/4+1/6+1/8... n terms
$\displaystyle \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots = \sum_{i=1}^{n} \frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2}\times \frac{1}{i}= \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i}$
this follows http://en.wikipedia.org/wiki/Harmoni...s_(mathematics)
As pickslides said, that is the same as $\displaystyle \frac{1}{2}\sum_{i=1}^n \frac{1}{i}$, the "harmonic series".
Since $\displaystyle \zeta(n)$ is defined as $\displaystyle \sum_{i=1}^\infty \frac{1}{i^n}$, $\displaystyle \sum_{i=1}^\infty \frac{1}{n}= \zeta(1)$ and so $\displaystyle \sum_{i=1}^\infty \frac{1}{2i}= \frac{1}{2}\zeta(a)$ but I do not know of any way to write the partial sums.