Value of $\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$ To find the value of:
$\displaystyle\sum_{n=2}^\infty (-1)^n ((\sum_{k=1}^\infty 1/{k^n})-1)$
The answer says: $1/2$
My attempt:
$=(\displaystyle\sum_{k=1}^\infty 1/{k^2})-1-(\sum_{k=1}^\infty 1/{k^3})+1+(\sum_{k=1}^\infty 1/{k^4})-1-\cdots$
$=(1+1/2^2+1/3^2+1/4^2+\cdots)-(1+1/2^3+1/3^3+1/4^3+\cdots)+(1+1/2^4+1/3^4+1/4^4+\cdots)-\cdots$
$=(1/2^2-1/2^3+1/2^4-\cdots)+(1/3^2-1/3^3+1/3^4-\cdots)+(1/4^2-1/4^3+1/4^4-...)+\cdots$
$=S_1+S_2+S_3+\cdots$
With:
$S_2=(1/2^2-1/2^3+1/2^4-\cdots)$
$S_3=(1/3^2-1/3^3+1/3^4-\cdots)$
$S_4=(1/4^2-1/4^3+1/4^4-\cdots)$
...
It looks like geometry series with alternating signs.
If I apply the geometric series value: $S=\frac{1}{1-z}$
Gives: 
$S_2=2-1-1/2=1/2$
$S_3=3/2-1-1/3=1/2-1/3$
$S_4=4/3-1-1/4=1/3-1/4$
...
Original equation:
$=1/2+1/2-1/3+1/3-1/4+1/4-1/5+\cdots=1$
This is different than the answer (1/2) given.
Could any one help? Thanks,

So I made calculation error when using the equation on geometric series.
$S_2=2/3-1+1/2=1/2-1/3$ (with z=-1/2)
$S_3=3/4-1+1/3=1/3-1/4$ (with z=-1/3)
$S_4=4/5-1+1/4=1/4-1/5$ (with z=-1/4)
Sum them up gives the value of original equation:
$=1/2-1/3+1/3-1/4+1/4-1/5+...=1/2$
