Can it be shown that,for all $m\in\mathbb{Z}\_+$ there exists at least one $k$ with respect to $m$ such that


$$1\le \frac{\sum\_{i=1}^{k-1}i^m}{k^m}<2$$


*Example*: let $m=1$ then $k=\{3,4\}$


This question has already been asked on MSE but it has not received a solution so that's why I posted here. Given claim solve my first claim from this post ["Observation on Erdős–Moser equation".](https://math.stackexchange.com/q/3546797/647719)


[Post on MSC](https://math.stackexchange.com/q/3552104/647719)



$\exists k$ s.t. $k^m\le 1^m+2^m+...+(k-1)^m <2\cdot k^m$?
Yes. Note that for large $k$ we have $1^m+\dots+(k-1)^m\geqslant k^m$. Choose minimal $k$ with such property. Then $k>2$ and $$2k^m>2(k-1)^m>1^m+\dots+(k-1)^m$$ 


by minimality.


