A well known identities of rare beauty is that:
$$\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac 1{\zeta(s)}$$
Where $\mu(n)$ is the Mobius function and $\zeta(s)$ is the Riemann Zeta function. So, in the context of summability theory/divergent series summation one can easily deduce that:
$$\sum_{n=1}^\infty \mu(n)=1-1-1+0-1+1-1+0+0+\dots=\frac 1{\zeta(0)}=-2$$
$$\sum_{n=1}^\infty \mu(n)n=1-2-3+0-5+6-7+0+0+\dots=\frac 1{\zeta(-1)}=-12$$
Or I'm not supposed to jump to that conclusion ? If I'm doing this correctly I can generalize the statement for general odd integers as:
$$\sum_{n=1}^\infty \mu(n)n^k=1-2^k-3^k+0-5^k+6^k-7^k+0+0+\dots=\frac 1{\zeta(-k)}=-\frac {k+1}{B_{k+1}}$$
Where $B_k$ are the Bernoulli numbers. However my problem is for even values of $k$ for which we have $\zeta(-k)=0$ and the situation became pretty messy. How can I solve this problem and for example assign in some way a finite value to:
$$\sum_{n=1}^\infty \mu(n)n^2=1-4-9+0-25+36-49+0+0+\dots=?$$