An inequality with fractional parts $$
\frac{n^k-n}{2}
\leq
\left\{\sqrt[k]{1}\right\} + \left\{\sqrt[k]{2}\right\} + \dots + \left\{\sqrt[k]{n^k}\right\} 
\leq
\frac{n^k-1}{2}
$$
how it can be proven?
 A: One side can be proved using concavity of $f(x) = x^{1/k}$: let's call
$$ S_m = \sum_{t=1}^{(m+1)^k-m^k-1}  \left \{ \sqrt[k]{m^k+t} \right \} $$
so that our sum $T$ is
$$ T = \sum_{m=1}^{n-1} S_m $$
i.e. we "split" the sum in the points where $\{ x^{1/k} \}$ is 0. Note that, by concavity:
$$ \sqrt[k]{m^k+t} = \left [ \left ( 1- \frac{t}{(m+1)^k-m^k} \right ) m^k + \left ( \frac{t}{(m+1)^k-m^k} \right ) (m+1)^k \right ]^{1/k} \ge \left ( 1- \frac{t}{(m+1)^k-m^k} \right )m + \left ( \frac{t}{(m+1)^k-m^k} \right ) (m+1) = m+ \frac{t}{(m+1)^k-m^k} $$
which implies, for $ t < (m+1)^k-m^k$, that 
$$ \left \{ \sqrt[k]{m^k+t} \right \} \ge \frac{t}{(m+1)^k-m^k} $$
Substituting it into $S_m$ yields:
$$ S_m = \sum_{t=1}^{(m+1)^k-m^k -1} \left \{ \sqrt[k]{m^k+t} \right \} \ge \sum_{t=1}^{(m+1)^k-m^k -1} \frac{t}{(m+1)^k-m^k} = \frac{(m+1)^k-m^k-1}{2} $$
In conclusion
$$ T = \sum_{m=1}^{n-1} S_m \ge \sum_{m=1}^{n-1} \frac{(m+1)^k-m^k-1}{2} = \frac{n^k-n}{2} $$
I think that even the other side can be proved by a similar "linear bound", maybe using the fact that the line through $( (m+1)^k, f( ..) ), ( (m+2)^k, f(..) )$ is always over the $f$ in the interval $ ] m^k, (m+1)^k [$. However, I can't prove it in this way!
Hope it helps, Andrea
A: Checking at wolfram it actually seems to be the opposite that $\frac{n^k-1}{2}$ is a lower bound and I will prove it below.
We can separate the sum into intervals like this by using that $\{n\}=n-\lfloor n\rfloor$ and the only integers are the $k$'th powers so they are equal to $0$. $$\sum_{d=2}^{2^k-1}(\sqrt[k]{d}-1)+\sum_{d=2^k+1}^{3^k-1}(\sqrt[k]{d}-2)+\sum_{d=3^k+1}^{4^k-1}(\sqrt[k]{d}-3)+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}(\sqrt[k]{d}-n+1)$$
Now notice we can write as $$\sum_{d=2}^{2^k-1}\sqrt[k]d+\sum_{d=2^k+1}^{3^k-1}\sqrt[k]d+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}\sqrt[k]d=\sum_{d=1}^{n^k}\sqrt[k]d-\sum_{d=1}^nd$$Now notice that
$$\sum_{d=2}^{2^k-1}(-1)+\sum_{d=2^k+1}^{3^k-1}(-2)+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}(-n+1)\\=(-1)(2^k-1-2)+(-2)(3^k-1-(2^k+1))+\cdots+(n-1)(n^k-1-((n-1)^k+1)\\=\sum_{d=1}^{n-1}d((d+1)^k-d^k-2)$$
Now combining that information we get that the sum is equal to
$$=\sum_{d=1}^{n^k}\sqrt[k]d-\sum_{d=1}^nd-\sum_{d=1}^{n-1}d((d+1)^k-d^k-2)\\=\sum_{d=1}^{n^k}\sqrt[k]d-\frac{n(n+1)}{2}+2\sum_{d=1}^{n-1}d-\sum_{d=1}^{n-1}d((d+1)^k-d^k)\\=\sum_{d=1}^{n^k}\sqrt[k]d-\frac{n(n+1)}{2}+n(n-1)-\sum_{d=1}^{n-1}d((d+1)^k-d^k)$$
Now by telescoping like method we get that
$$\sum_{d=1}^{n-1}d((d+1)^k-d^k)=2^k-1+2\cdot 3^k-2\cdot 2^k+3\cdot 4^k-3\cdot 3^k+\cdots + (n-1)\cdot n^k-(n-1)\cdot (n-1)^k=(n-1)n^k-((n-1)^k+(n-2)^k+\cdots + 2^k+1)\\=n^{k+1}-\sum_{d=1}^nd^k$$ So we are left with $$\sum_{d=1}^{n^k}\sqrt[k]{d}+\frac{n(n-3)}{2}-\left(n^{k+1}-\sum_{d=1}^nd^k\right)\\=\sum_{d=1}^{n^k}\sqrt[k]d+\frac{(n-2)(n-1)}{2}-n^{k+1}+\sum_{d=1}^nd^k$$
Now lets say $n> 3$ and $k>3$ to avoid the edge cases. For the first inequality I used the formulas from Asymptotic behaviour of sums of consecutive powers $$\sum_{d=1}^nd^k>  \frac{n^{k+1}}{k+1}+\frac{n^k}2$$
Also  $$\sum_{d=1}^{n^k}\sqrt[k]{d}\geq\int_0^{n^k}\sqrt[k]xdx=\frac{kn^{k+1}}{k+1}$$
Now adding those estimates ignoring $(\frac{(n-1)(n-2)}{2})$ since it's positive we get
$$\left\{\sqrt[k]{1}\right\} + \left\{\sqrt[k]{2}\right\} + \dots + \left\{\sqrt[k]{n^k}\right\}=\sum_{d=1}^{n^k}\sqrt[k]d+\frac{(n-2)(n-1)}{2}-n^{k+1}+\sum_{d=1}^nd^k \\>\frac{kn^{k+1}}{k+1}+\frac{n^{k+1}}{k+1}+\frac{n^k}2-n^{k+1}=\frac{n^k}{2}> \frac{n^k-1}{2}$$
A: One can show, by Bernoulli's Inequality and induction, that $$\frac{n^{\alpha+1}}{\alpha+1}<\sum_{k=1}^n k^\alpha$$ when $\alpha>0$. Applying this to $``n"=n^k$ and $\alpha=1/k$ we actually get your sum is bounded below by the quantity
$$\frac{n^{k+1}}{\frac{1}{k}+1}\,.$$
Both of your bounds lie below this quantity.
