# $1^\alpha+2^\alpha+3^\alpha+\cdots+n^\alpha$

Let $$\alpha>0$$ and $$m$$ be a positive integers, use Euler's summation formula we can prove that there exists a constant $$C$$ such that $$\sum_{k=1}^nn^\alpha=\frac{n^{\alpha+1}}{\alpha+1}+\frac{n^\alpha}{2}+{\color{red}C} +\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}\alpha(\alpha-1)\cdots(\alpha-2k+2)n^{\alpha-2k+1}$$ $$+\frac{\alpha(\alpha-1)\cdots(\alpha-2m)}{(2m+1)!} \int_1^nB_{2m+1}(x-[x])\cdot x^{\alpha-2m-1}dx.$$

$$=\frac{n^{\alpha+1}}{\alpha+1}+\frac{n^\alpha}{2}+{\color{red}{C_1}} +\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}\alpha(\alpha-1)\cdots(\alpha-2k+2)n^{\alpha-2k+1}+O(n^{\alpha-2m-1}).$$

Here the two constants $$C$$ and $$C_1$$ maybe different.

My question is: can we prove that the constant $$C_1$$ equals to $$\zeta(-\alpha)$$? Here $$\zeta(s)$$ is the Riemann zeta function.

Let $$f(x)$$ be a smooth function over $$(0,+\infty)$$, by Euler–Maclaurin formula, we have that $$f(1)+f(2)+\cdots+f(n)=F(n)+C+\frac{f(n)}{2}+\sum_{k=1}^m\frac{B_{2k}}{(2k)!}f^{(2k-1)}(n)$$ $$-\int_n^\infty\frac{B_{2m+1}(\{x\})}{(2m+1)!}f^{(2m+1)}(x)dx,$$ where $$C=\frac{f(1)}{2}-\left(F(1)+\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}f^{(2k-1)}(1)\right)$$ $$+\int_1^\infty\frac{B_{2m+1}(\{x\})}{(2m+1)!}f^{(2m+1)}(x)dx..$$ Here $$F(x)$$ is a primitive function of $$f(x)$$. Let $$f(x)=x^\alpha$$ and $$F(x)=\frac{x^{\alpha+1}}{\alpha+1}$$, then $$C=\frac{1}{2}-\frac{1}{\alpha+1}-\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}\alpha(\alpha-1)\cdots(\alpha-2k+2)$$ $$+\frac{\alpha(\alpha-1)\cdots(\alpha-2m)}{(2m+1)!} \int_1^\infty B_{2m+1}(x-[x])\cdot x^{\alpha-2m-1}dx.$$
For $$\Re s>1$$, Let $$\zeta(s)=\sum\limits_{n=1}^\infty\dfrac{1}{n^s}$$ be the Riemann zeta function. Similarly, by Euler-Maclaurin formula, we have that $$\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\sum_{k=1}^m\frac{s(s+1)\cdots(s+2k-2)}{(2k)!}B_{2k}- \frac{s(s+1)\cdots(s+2m)}{(2m+1)!}\int_1^\infty\frac{B_{2m+1}(x-[x])}{x^{s+2m+1}}dx$$ and use this formula we can extend the definition of $$\zeta(s)$$ to the half plane $$\Re s>-2m$$. Take $$s=-\alpha$$, we have that $$\zeta(-\alpha)=\frac{1}{2}-\frac{1}{\alpha+1}-\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}\alpha(\alpha-1)\cdots(\alpha-2k+2)$$ $$+\frac{\alpha(\alpha-1)\cdots(\alpha-2m)}{(2m+1)!} \int_1^\infty\frac{B_{2m+1}(x-[x])}{x^{-\alpha+2m+1}}dx.$$ Hence $$C=\zeta(-\alpha)$$ and thus $$1^\alpha+2^\alpha+\cdots+n^\alpha=\frac{n^{\alpha+1}}{\alpha+1}+\frac{n^\alpha}{2}+\zeta(-\alpha) +\sum_{k=1}^{m}\frac{B_{2k}}{(2k)!}\alpha(\alpha-1)\cdots(\alpha-2k+2)n^{\alpha-2k+1}+O(n^{\alpha-2m-1})~(\alpha<2m).$$