extension of $\zeta(s)=\lim_{n\to\infty}\sum_{k=1}^n 1/k^s -\int_0^n 1/x^s$

Question

I saw $\zeta(1/2)=\lim_{n\to\infty}\sum_{k=1}^n 1/\sqrt k-\int_0^n\ 1/\sqrt x\ dx$ here and similarly $\zeta(s)=\lim_{n\to\infty}\sum_{k=1}^n 1/k^s -\int_0^n 1/x^s$ for $0<s<1$.

Rewriting this as $\zeta(s) = \lim_{n\to\infty}\sum_{k=1}^n 1/k^s -\int_1^n 1/x^s + 1/(s-1)$, I observed it also holds for $s > 1$. (It just follows from $\zeta$'s definition.)
(For example, $\zeta(1/3) = \lim_{n\to\infty}\sum_{k=1}^n 1/k^3 -\int_1^n 1/x^{1/3} - 3/2$, $\zeta(3) = \lim_{n\to\infty}\sum_{k=1}^n 1/k^3 -\int_1^n 1/x^3 + 1/2$.)
For $s=1$, I can interpret it as $\lim_{s\to 1} \zeta(s)-1/(s-1)=\lim_{n\to\infty}\sum_{k=1}^n 1/k -\int_1^n 1/x$, which is the Euler's constant.

Now I wonder is there some interpretation to apply this for $s\leq 0$?
(Some modification is needed because for example
$\zeta(-1) =-1/12 \neq \lim_{n\to\infty}\sum_{k=1}^n k -\int_1^n x + 1/2 = \lim_{n\to\infty} n/2+1/2 $.
but I could not find any natural modification.)

Answer 1

Suppose that $\Re s>1$. In terms of the Hurwitz zeta function (http://dlmf.nist.gov/25.11) $$ \sum\limits_{k = 1}^n {\frac{1}{{k^s }}} = \zeta (s) - \sum\limits_{k = n + 1}^\infty {\frac{1}{{k^s }}} = \zeta (s) + \frac{1}{{n^s }} - \sum\limits_{k = n}^\infty {\frac{1}{{k^s }}} = \zeta (s) + \frac{1}{{n^s }} - \zeta (s,n), $$ i.e., $$ \zeta (s) = \sum\limits_{k = 1}^n {\frac{1}{{k^s }}} - \frac{1}{{n^s }} + \zeta (s,n). $$ By analytic continuation, this formula is valid for all complex $s\neq 1$. Now you can use, for example, http://dlmf.nist.gov/25.11.E28 to obtain a more explicit expression that is valid in half-planes of the form $\Re s>-(2N+1)$. You can also use the asymptotic expansion $$ \zeta (s,n) \sim \frac{{n^{1 - s} }}{{s - 1}} + \frac{1}{{2n^s }} + \sum\limits_{m = 1}^\infty {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (s + 2m - 1)}}{{\Gamma (s)}}\frac{1}{{n^{s + 2m - 1} }}} $$ as $n\to +\infty$ with fixed $s \neq 1$ (http://dlmf.nist.gov/25.11.E43). Here $B_m$ denotes the Bernoulli numbers. In particular, $$ \zeta (s) = \mathop {\lim }\limits_{n \to + \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{k^s }}} + \frac{{n^{1 - s} }}{{s - 1}} - \frac{1}{{2n^s }} + \sum\limits_{m = 1}^N {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (s + 2m - 1)}}{{\Gamma (s)}}\frac{1}{{n^{s + 2m - 1} }}} } \right) $$ for all $s\neq 1$ and $N$ satisfying $\Re s > - (2N + 1)$. Note that $$ \frac{{n^{1 - s} }}{{s - 1}} = - \int_1^n {\frac{{dx}}{{x^s }}} + \frac{1}{{s - 1}} . $$