Question

`( \zeta(2))/(2) - \frac{ \zeta(3)}3 + ( \zeta(4))/(4) - ( \zeta(5))/(5) + \cdots \cdots \\`​

Answer 1

Recall the Weierstrass product definition of the gamma function,

`\displaystyle \frac1{\Gamma(z)} = ze^(\gamma z) \prod_(n=1)^\infty \left(1+\frac zn\right) e^(-z/n)`

from which we can obtain the log-gamma (i.e. logarithm of gamma) function,

`\displaystyle -\ln\Gamma(z) = \ln(z) + \gamma z + \sum_(n=1)^\infty \left(\ln\left(1 + \frac zn\right) - \frac zn\right)`

where
`\gamma` is the Euler-Mascheroni constant.

Let
`z=1`, so that

`\displaystyle \gamma = \sum_(n=1)^\infty \left(\frac1n - \ln\left(1 + \frac 1n\right)\right)`

Recall the power series

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

Along with the definition of the Riemann zeta function,

`\displaystyle \zeta(a) = \sum_(b=1)^\infty \frac1{b^a}`

we find that

`\displaystyle \gamma = \sum_(n=1)^\infty \left(\frac1n + \sum_(m=1)^\infty \frac1m\left(-\frac1n\right)^m\right) \\\\ ~~~~ = \sum_(n=1)^\infty \sum_(m=2)^\infty \frac1m \left(-\frac1n\right)^m \\\\ ~~~~ = \sum_(m=2)^\infty \frac{(-1)^m}m \sum_(n=1)^\infty \frac1{n^m} \\\\ ~~~~ = \sum_(m=2)^\infty \frac{(-1)^m \zeta(m)}m`

so the original sum's value is
`\boxed{\gamma\approx0.577216}`.