TITLE: Riemann zeta for real argument between 0 and 1 using Mellin, with short asymptotic expansion
QUESTION [8 upvotes]: The following would appear to be true. 
For real $0 < \sigma < 1,$ we seem to have a very satisfying sum minus integral limit,
$$  \zeta(\sigma)  \; \; = \; \;  \lim_{n \rightarrow \infty} \; \; \left( \sum_{k=1}^n \frac{1}{k^\sigma}   \right) - \frac{n^{1-\sigma}}{1-\sigma} \; \; . $$
A little more detail in these examples: 
$$ \lim_{n \rightarrow \infty} \; \; \left( \sum_{k=1}^n \frac{1}{\sqrt k}   \right) - 2 \sqrt n - \frac{1}{2 \sqrt n} + \frac{1}{24 n^{3/2}} \; \; = \; \; \zeta \left(\frac{1}{2}\right) \approx -1.460354508809586  $$
$$ \lim_{n \rightarrow \infty} \; \; \left( \sum_{k=1}^n \frac{1}{\sqrt[3] k}   \right) - \frac{3 n^{2/3}}{2} - \frac{1}{2 \sqrt[3] n} + \frac{1}{36 n^{4/3}} \; \; = \; \; \zeta \left(\frac{1}{3}\right) \approx -0.973360248350782  $$
So, are the above items really really true (rather than wishful thinking), and, if so, given $0 < \sigma < 1,$ what real numbers $A = A(\sigma), B = B(\sigma),$  make the short asymptotic expansion below correct?
$$ \left( \sum_{k=1}^n \frac{1}{k^\sigma}   \right) - \frac{n^{1-\sigma}}{1-\sigma} - A n^{-\sigma} + B n^{-1-\sigma} \; \; = \; \;  \zeta(\sigma) \; \; + \; \;  O(n^{-2-\sigma}) \; \; ? $$
Here is an estimate of $\zeta\left( \frac{1}{5} \right)$ using Daniel's expansion as far as the $3+\sigma$ term, ignoring $5+\sigma$: $$ \zeta(\sigma) =\left( \sum_{k=1}^n \frac{1}{k^\sigma}   \right) - \frac{n^{1-\sigma}}{1-\sigma} - \frac{1}{2 n^\sigma} + \frac{\sigma}{12 n^{1 + \sigma}} - \frac{\sigma (1 + \sigma)(2 + \sigma)}{720 n^{3+\sigma}} + O \left( \frac{1}{n^{5 + \sigma}} \right)$$
I think it will let me fit 33 lines in a "code" window without introducing a scroll bar.
    1   1                   -0.7340666666666666   
    2   1.870550563296124   -0.7339263390330826   
    3   2.673292125056355   -0.7339216399387463   
    4   3.431150408311554   -0.7339210905379737   
    5   4.155930071989249   -0.7339209776636304   
    6   4.854757190760828   -0.7339209455319796   
    7   5.532368104161309   -0.7339209342063361   
    8   6.192122059547756   -0.7339209295629763   
    9   6.836516074525011   -0.7339209274322004   
   10   7.467473419005204   -0.7339209263652269   
   11   8.086517339689049   -0.7339209257924113   
   12   8.694881681582254   -0.7339209254668952   
   13   9.293584537123753   -0.7339209252729478   
   14   9.883479099480143   -0.7339209251526775   
   15   10.46528985863283   -0.7339209250754953   
   16   11.03963903613135   -0.7339209250244763   
   17   11.60706632180293   -0.7339209249898626   
   18   12.16804389452603   -0.7339209249658375   
   19   12.72298804735438   -0.7339209249488197   
   20   13.27226831900744   -0.7339209249365428   
   21   13.81621476189632   -0.7339209249275397   
   22   14.35512379575268   -0.7339209249208374   
   23   14.8892629725969    -0.7339209249157863   
   24   15.41887489312131   -0.7339209249119263   
   25   15.94418045400206   -0.733920924908948   
   26   16.46538156214071   -0.7339209249066238   
   27   16.98266342011249   -0.7339209249047942   
   28   17.49619646365717   -0.7339209249033369   
   29   18.00613801451264   -0.7339209249021712   
   30   18.51263369862476   -0.7339209249012291   
   31   19.01581866962328   -0.733920924900462   
   32   19.51581866962328   -0.7339209248998368   
   33   20.0127509533112    -0.7339209248993227

REPLY [5 votes]: There is  a standard technique  that produces the  complete asymptotic
expansion  for this  sum and  many  others like  it, which  is to  use
harmonic sums and Mellin transforms.
Introduce the telescoping sum where $0\lt\sigma\lt 1$
$$S(x) = \sum_{k\ge 1} \left(\frac{1}{k^\sigma}-
\frac{1}{(x+k)^\sigma}\right).$$
This sum has the property that 
$$S(n) = \sum_{p=1}^n \frac{1}{p^\sigma},$$ 
so that $S(n)$ is the value we are looking for.
Re-write the sum as follows:
$$S(x) = \sum_{k\ge 1} \frac{1}{k^\sigma}
 \left(1-\frac{1}{(x/k+1)^\sigma}\right).$$
The sum term is harmonic and  may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^\sigma}, 
\quad \mu_k = \frac{1}{k} \quad \text{and} \quad
g(x) = 1 - \frac{1}{(1+x)^\sigma}.$$
It follows that
$$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1} \frac{1}{k^\sigma} \times k^s
= \zeta(\sigma-s)$$
which has fundamental strip $\sigma-s > 1$ or $s < \sigma-1.$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \left(1 - \frac{1}{(1+x)^\sigma}\right) x^{s-1} dx$$
which is immediately seen to be a beta function integral with value
$$g^*(s) = - \frac{1}{\Gamma(\sigma)} \Gamma(s)\Gamma(\sigma-s)$$
and  fundamental strip  $\langle -1,  0  \rangle.$ We  check that  the
abscissa of  convergence of the  zeta function  term is right  in this
strip as required. It follows that  the Mellin transform $Q(s)$ of the
harmonic sum $S(x)$ is given by
$$Q(s) =
- \frac{1}{\Gamma(\sigma)} \Gamma(s)\Gamma(\sigma-s)
\zeta(\sigma-s).$$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} 
\int_{\sigma-1-\varepsilon-i\infty}
^{\sigma-1-\varepsilon+i\infty} Q(s)/x^s ds$$
which we  evaluate by  shifting it  to the right  for an  expansion at
infinity.
First  treat the  pole from  the zeta  function term  at $s=\sigma-1$,
which has
$$\mathrm{Res}(Q(s)/x^s; s=\sigma-1) =
-\frac{1}{\Gamma(\sigma)} \Gamma(\sigma-1)\Gamma(1)
\times -1 \times x^{1-\sigma}
= -\frac{1}{1-\sigma} x^{1-\sigma}.$$
For the pole at $s=0$ from the simple gamma function term we obtain
$$\mathrm{Res}(Q(s)/x^s; s=0) = 
-\frac{1}{\Gamma(\sigma)} \Gamma(\sigma) \zeta(\sigma)
= -\zeta(\sigma).$$
For the pole at $s=\sigma$ from the compound gamma function term we obtain
$$\mathrm{Res}(Q(s)/x^s; s=\sigma) =
 -\frac{1}{\Gamma(\sigma)} \Gamma(\sigma) \times -1 \times\zeta(0)
\times \frac{1}{x^\sigma}
= -\frac{1}{2} \frac{1}{x^\sigma}.$$
The remaining poles are at $s = q+\sigma$ where $q\ge 1$ and contribute
$$\mathrm{Res}(Q(s)/x^s; s=q+\sigma) =
-\frac{1}{\Gamma(\sigma)} \Gamma(\sigma+q) 
\frac{(-1)^{q+1}}{q!} \zeta(-q) \frac{1}{x^{q+\sigma}}
\\ = - \prod_{p=0}^{q-1} (p+\sigma) \times
 \frac{(-1)^{q+1}}{q!} (-1)^q \frac{B_{q+1}}{q+1} \frac{1}{x^{q+\sigma}}
= {q+\sigma-1\choose q} 
\frac{B_{q+1}}{q+1}  \frac{1}{x^{q+\sigma}}.$$
The  zero   values  of  the  Bernoulli   numbers  correctly  represent
cancelation of  the gamma function poles  by the trivial zeros  of the
zeta function.
Setting $x=n$  and observing that  the shift  to the right  produces a
minus sign we obtain the following asymptotic expansion:
$$S(n) = \sum_{p=1}^n \frac{1}{p^\sigma}
\\ \sim \frac{1}{1-\sigma} n^{1-\sigma}
+ \zeta(\sigma) + \frac{1}{2} \frac{1}{n^\sigma}
- \sum_{q\ge 1} {q+\sigma-1\choose q}
\frac{B_{q+1}}{q+1}  \frac{1}{n^{q+\sigma}}.$$
Actually computing the Bernoulli number terms we get for the example
by OP with $\sigma=1/3$ the expansion
$$3/2\,{n}^{2/3}+\zeta  \left( 1/3 \right) 
+1/2\,{\frac {1}{\sqrt [3]{n}}}
-1/36\,{n}^{-4/3}+{\frac {7}{4860\,{n}^{10/3}}}
\\-{\frac {13}{26244\,{n}^{16/3}}}
+{\frac {247}{590490}{n}^{-{\frac {22}{3}}}}
-{\frac {6175}{9565938}{n}^{-{\frac {28}{3}}}}
\\+{\frac {406999}{258280326}{n}^{-{\frac {34}{3}}}}
-{\frac {12966835}{2324522934}{n}^{-{\frac {40}{3}}}}+\cdots$$
This    MSE    link
points to a series of similar calculations.