Generalisation of $ \sum \frac {1}{k}-\ln n=\gamma$ to $0 \lt\alpha \lt1 , \sum \frac{1}{k^\alpha}-f(n)= \beta$ looking at Find the value of : $\lim\limits_{n\rightarrow\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac{1}{\sqrt k}\right)$
and knowing that for 
$$\alpha=1 ,\lim_{ n \to \infty}  \sum_{k=1}^n \frac {1}{k^\alpha}-\ln n=\gamma$$ 
Makes one wonder if there are other results for 
 $0 \lt \alpha \lt 1 $
$$\lim_{ n \to \infty}  \sum_{k=1}^n \frac {1}{k^\alpha}-f(n)=\beta$$
where $f,\beta$ are determined by the value of $\alpha$
is there a topic that relates to some results similar to above? 
 A: Suppose that $\Re \alpha>1$. In terms of the Hurwitz zeta function (http://dlmf.nist.gov/25.11)
$$
\sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) - \sum\limits_{k = n + 1}^\infty  {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) + \frac{1}{{n^\alpha }} - \sum\limits_{k = n}^\infty  {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) + \frac{1}{{n^\alpha }} - \zeta (\alpha,n),
$$
i.e.,
$$
\zeta (\alpha) = \sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  - \frac{1}{{n^\alpha }} + \zeta (\alpha,n).
$$
By analytic continuation, this formula is valid for all complex $\alpha\neq 1$. 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 $\Re \alpha>-(2N+1)$. You can also use the asymptotic expansion
$$
\zeta (\alpha,n) \sim \frac{{n^{1 - \alpha} }}{{\alpha - 1}} + \frac{1}{{2n^\alpha }} + \sum\limits_{m = 1}^\infty  {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}} 
$$
as $n\to +\infty$ with fixed $\alpha \neq 1$ (http://dlmf.nist.gov/25.11.E43). Here $B_m$ denotes the Bernoulli numbers. In particular,
$$
\zeta (\alpha) = \mathop {\lim }\limits_{n \to  + \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  + \frac{{n^{1 - \alpha} }}{{\alpha - 1}} - \frac{1}{{2n^\alpha }} + \sum\limits_{m = 1}^N {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}} } \right)
$$
for all $\alpha\neq 1$ and $N$ satisfying $\Re \alpha >  - (2N + 1)$. Therefore, with your notation, $\beta =\zeta(\alpha)$ and
$$
f(n)= -\frac{{n^{1 - \alpha} }}{{\alpha - 1}} + \frac{1}{{2n^\alpha }} - \sum\limits_{m = 1}^N {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}}
$$
with $N = \left\lceil { - \frac{{\Re \alpha  + 1}}{2}} \right\rceil$, $\alpha\neq 1$ and with empty sums being interpreted as $0$.
A: You could use
$$\zeta(\alpha)=\frac 1 {\alpha-1}+\sum_{n=0}^\infty (-1)^n\,\frac{ \gamma _n}{n!} \,(\alpha-1)^n$$ which is very accurate (just try summing from $0$ to $4$).
