Finding the value of limit when $k\in(-1,1)$ My question arises from How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$.
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1.If $k\in(1,+\infty),$  then $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=\frac{k}{12};$$
2. If $k=1,$ then $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=0;$$
3.If $k\in(-\infty,-1),$  then 
 $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=+\infty.$$
When $k\in (-1,1),$ I guess we could get $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=-\infty.$$
Euler-Maclaurin Summation Formula is at work in this case,but I need more convenient method to solve the problem.
 Can anyone give me any hints on how to start it? Any help will be appreciated.
 A: For $k \in (-1,0]$, we have a very simple argument.
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} = \frac{1}{n}\sum_{m = 1}^n \biggl(\frac{m}{n}\biggr)^k$$
is a Riemann sum for the integral $\int_0^1 x^k\,dx$ for the equidistant partition $0 < \frac{1}{n} < \dotsc < \frac{n-1}{n} < 1$ with the right endpoints of the partition intervals chosen as the points where the integrand is evaluated. Since $x \mapsto x^k$ is non-increasing for $k \leqslant 0$, it follows that
$$\frac{1}{n}\sum_{m = 1}^n \biggl(\frac{m}{n}\biggr)^k \leqslant \int_0^1 x^k\,dx = \frac{1}{k+1}$$
(with strict inequality for $k < 0$). Hence
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n} \leqslant -\frac{1}{2n}$$
and
$$n^2\biggl(\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n}\biggr) \leqslant -\frac{n}{2},$$
from which we immediately see that the limit is $-\infty$.
For $k \in (0,1)$, the argument is a bit more complex. We first note that
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{2n} = \frac{1}{n}\sum_{m = 1}^n \frac{1}{2}\Biggl(\biggl(\frac{m-1}{n}\biggr)^k + \biggl(\frac{m}{n}\biggr)^k\Biggr)$$
is a trapezium sum for the integral $\int_0^1 x^k\,dx$.
Next we note that for a twice continuously differentiable $f \colon [a,b] \to \mathbb{R}$, with $\mu := \frac{a+b}{2}$, we have
\begin{align}
\frac{1}{2} \int_a^b (x-a)(b-x)f''(x)\,dx
&= \frac{1}{2}(x-a)(b-x)f'(x)\biggr\rvert_a^b + \int_a^b (x-\mu)f'(x)\,dx \\
&= \int_a^b (x-\mu)f'(x)\,dx \\
&= (x-\mu)f(x)\biggr\rvert_a^b - \int_a^b f(x)\,dx \\
&= \frac{b-a}{2}\bigl(f(a) + f(b)\bigr) - \int_a^b f(x)\,dx,
\end{align}
and by the mean value theorem of integration,
$$\frac{1}{2} \int_a^b (x-a)(b-x)f''(x)\,dx = \frac{f''(\xi)}{2}\int_a^b (x-a)(b-x)\,dx = \frac{(b-a)^3}{12} f''(\xi)$$
for some $\xi \in (a,b)$. Although $f \colon x \mapsto x^k$ is for $k \in (0,1)$ not differentiable at $0$, these formulae remain correct also with $a = 0$ for that function. Thus
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n} = \frac{k(k-1)}{12 n^3}\sum_{m = 1}^n \biggl(\frac{\eta_{n,m}}{n}\biggr)^{k-2}$$
for some $\eta_{n,m} \in (m-1,m)$. Then
$$n^2\biggl(\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n}\biggr) = n^{1-k}\frac{k(k-1)}{12} \sum_{m = 1}^n \frac{1}{\eta_{n,m}^{2-k}} \to -\infty$$
since $1-k > 0$, $\frac{k(k-1)}{12} < 0$, and
$$\sum_{m = 1}^n \frac{1}{\eta_{n,m}^{2-k}} \geqslant \frac{1}{\eta_{n,1}^{2-k}} > 1$$
for all $n$.
