Recall the Weierstrass product definition of the gamma function,
[tex]\displaystyle \frac1{\Gamma(z)} = ze^{\gamma z} \prod_{n=1}^\infty \left(1+\frac zn\right) e^{-z/n}[/tex]
from which we can obtain the log-gamma (i.e. logarithm of gamma) function,
[tex]\displaystyle -\ln\Gamma(z) = \ln(z) + \gamma z + \sum_{n=1}^\infty \left(\ln\left(1 + \frac zn\right) - \frac zn\right)[/tex]
where [tex]\gamma[/tex] is the Euler-Mascheroni constant.
Let [tex]z=1[/tex], so that
[tex]\displaystyle \gamma = \sum_{n=1}^\infty \left(\frac1n - \ln\left(1 + \frac 1n\right)\right)[/tex]
Recall the power series
[tex]\displaystyle \ln(1 + x) = - \sum_{n=1}^\infty \frac{(-x)^n}n[/tex]
Along with the definition of the Riemann zeta function,
[tex]\displaystyle \zeta(a) = \sum_{b=1}^\infty \frac1{b^a}[/tex]
we find that
[tex]\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[/tex]
so the original sum's value is [tex]\boxed{\gamma\approx0.577216}[/tex].
