Question

`\rm \sum_(n = 0)^ \infty \frac{( - 1) {}^(n) \zeta(n) }{ {2}^(n) } \\`​

Answer 1

Assuming the correction I suggested in comments, our sum is

`\displaystyle S = \sum_(n=2)^\infty ((-1)^n \zeta(n))/(2^n)`

Rewrite with the definition of the Riemann zeta function, and condense the subsequent double series.

`\displaystyle S = \sum_(n=2)^\infty ((-1)^n)/(2^n) \sum_(m=1)^\infty \frac1{m^n} \\\\ ~~~~ = \sum_(n=2)^\infty \sum_(m=1)^\infty \left(-\frac1{2m}\right)^n`

The condition for Fubini's theorem is met so we can interchange the sums.

`\displaystyle S = \sum_(m=1)^\infty \sum_(n=2)^\infty \left(-\frac1{2m}\right)^n`

Sum the geometric series.

`\displaystyle S = \sum_(m=1)^\infty \left(\sum_(n=0)^\infty \left(-\frac1{2m}\right) - 1 + \frac1{2m}\right) \\\\ ~~~~ = \sum_(m=1)^\infty \left(\frac1{1+\frac1{2m}} - 1 + \frac1{2m}\right) \\\\ ~~~~ = \sum_(m=1)^\infty \left(\frac1{2m} - \frac1{2m+1}\right)`

Considering the terms in the summand as even/odd parts, we can condense this further to the alternating series

`\displaystyle S = \sum_(m=2)^\infty \frac{(-1)^m}m`

and recalling the power series

`\displaystyle \ln(1+x) = - \sum_(n=1)^\infty \frac{(-x)^n}n`

we let
`x=1`; it follows that

`\displaystyle S = \sum_(m=1)^\infty \frac{(-1)^m}m + 1 = \boxed{1 - \ln(2)}`