convergence to a generalized Euler constant and relation to Zeta serie Let $0 \leq a \leq 1$ be a real number. I would like to know how to prove that the following sequence converges:
$$u_n(a)=\sum_{k=1}^n k^a- n^a \left(\frac{n}{1+a}+\frac{1}{2}\right)$$
For $a=1$:
$$u_n(1)=\sum\limits_{k=1}^{n} k- n \left(\frac{n}{1+1}+\frac{1}{2}\right)= \frac{n(n+1)}{2}-\frac{n(n+1)}{2}=0$$
so  $u_n(1)$ converges to $0$.
for $a=0$: 
$$u_n(0)=\sum\limits_{k=1}^{n} 1-  \left(\frac{n}{1+0}+\frac{1}{2}\right) = n-n+\frac{1}{2}=\frac{1}{2}$$
so  $u_n(0)$ converges to $1/2$.
In genaral, the only idea I have in mind is the Cauchy integral  criterion but it does not work because $k^a$ is an increasing function,
Do the proof involves Zeta serie ?
 A: To avoid   a black box formula
Since the function $x\mapsto x^a$ is increasing then we find easily the asymptotic expansion
$$\sum_{k=1}^n k^a\sim_\infty \int_1^n x^a dx\sim_\infty \frac{1}{a+1}n^{a+1}\tag{AE}$$
hence we have
$$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+O\left(n^{a}\right)\tag{1}$$
Now let's improve the equality $(1)$ so let
$$v_n=\sum_{k=1}^n k^a-\frac{1}{a+1}n^{a+1}$$
so we have
$$v_{n+1}-v_n=(n+1)^a-\frac{1}{a+1}(n+1)^{a+1}+\frac{1}{a+1}n^{a+1}$$
then we expand using the binomial formula we find
$$v_{n+1}-v_n\sim_\infty  \frac{a}{2}n^{a-1}$$
and by telescoping and repeating the same method as in $(AE)$ we have:
$$v_n\sim_\infty \frac{a}{2}\int_1^n x^{a-1}dx\sim_\infty \frac{n^a}{2}$$
hence finaly we find 
$$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+ \frac{n^a}{2}+O\left(n^{a-1}\right)$$
and we conclude.
A: From this answer you have an asymptotics
$$
\sum_{k=1}^n k^a = \frac{n^{a+1}}{a+1} + \frac{n^a}{2} + \frac{a n^{a-1}}{12} +  O(n^{a-3})
$$
Use it to prove that $u_n(a)$ converges.
