Can we use analytic continuation to obtain $\sum_{n=1}^\infty n = b, b\neq -\frac{1}{12}$

Let $f_n(z,a)=\frac n{(n+a)^z}$. Taking the analytic continuation as $z\to0$ on the sum of $f$, we have

$$\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=\frac{a^2}2-\frac1{12}$$

The proof is not so bad. Notice that

$$\frac\partial{\partial x}\frac1{(xn+a)^{z-1}}=\frac n{(xn+a)^z}$$

And further that

$$\sum_{n=1}^\infty\frac1{(xn+a)^{z-1}}=\frac1{x^{z-1}}\sum_{n=1}^\infty\frac1{(n+\frac ax)^{z-1}}=\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}$$

where we use the Hurwitz zeta function. Differentiating with respect to $x$ then gives

$$\frac\partial{\partial x}\frac{\zeta(z-1,1+\frac ax)}{x^{z-1}}=\frac{-ax^{z-3}\zeta(z,1+\frac ax)-(z-1)x^{z-2}\zeta(z-1,1+\frac ax)}{x^{2z-2}}$$

Let $x=1$ and you end up with

$$\sum_{n=1}^\infty f_n(z,a)=-a\zeta(z,1+a)-(z-1)\zeta(z-1,1+a)$$

And as $z\to0^+$, we get...

$$\lim_{z\to0^+}f_n(z,a)=n\\\lim_{z\to0^+}\sum_{n=1}^\infty f_n(z,a)=-a^2\zeta(0)+\zeta(-1)=\frac{a^2}2-\frac1{12}$$

Let $f_n(s) = n^{-s}+ (s+1) e^{-(s+1)n} (b-\zeta(-1))$. Then $f_n(-1) = n$ and for $Re(s) > 1$ : $$F(s) = \sum_{n=1}^\infty f_n(s) = \zeta(s)+(b-\zeta(-1))\frac{s+1}{e^{s+1}-1}$$ It can be continued analytically to the complex plane minus $s=1$ and $s=-1+2ik \pi, k \in \mathbb{Z}^*$ : $$F(-1) = \zeta(-1) + (b-\zeta(-1))\lim_{s \to -1}\frac{s+1}{e^{s+1}-1}=b$$