Question Number 153392 by ArielVyny last updated on 06/Sep/21
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{a}} =?\:\:\: \\ $$
Commented by prakash jain last updated on 07/Sep/21
$$\zeta\left(−{a}\right)−\zeta\left(−{a},{n}+\mathrm{1}\right) \\ $$$$\mathrm{i}\:\mathrm{dont}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form} \\ $$$$\mathrm{formula}. \\ $$$$\zeta−\:\mathrm{Riemann}\:\mathrm{zeta}\:\mathrm{function} \\ $$