1. ## sequence series

sum of

1/2+1/4+1/6+1/8... n terms

2. 1/2 + 1/4 + 1/8 + ...
= (1/2)/(1 - 1/2)
= 1

This is sum to infinity of a geometric progression, with a = 1/2, r = 1/2
S(infinity) = a/(1 - r)

3. $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots = \sum_{i=1}^{n} \frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2}\times \frac{1}{i}= \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i}$

this follows http://en.wikipedia.org/wiki/Harmoni...s_(mathematics)

4. Originally Posted by gohpihan
1/2 + 1/4 + 1/8 + ...
= (1/2)/(1 - 1/2)
= 1

This is sum to infinity of a geometric progression, with a = 1/2, r = 1/2
S(infinity) = a/(1 - r)
this maybe true but is not what the OP asked.

5. Originally Posted by pickslides
$\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots = \sum_{i=1}^{n} \frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2}\times \frac{1}{i}= \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i}$

this follows http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

i have got the solution graphically as 1.2 but how to do it with zeta function

6. As pickslides said, that is the same as $\frac{1}{2}\sum_{i=1}^n \frac{1}{i}$, the "harmonic series".
Since $\zeta(n)$ is defined as $\sum_{i=1}^\infty \frac{1}{i^n}$, $\sum_{i=1}^\infty \frac{1}{n}= \zeta(1)$ and so $\sum_{i=1}^\infty \frac{1}{2i}= \frac{1}{2}\zeta(a)$ but I do not know of any way to write the partial sums.