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Question Number 152900 by SANOGO last updated on 03/Sep/21

Answered by mindispower last updated on 03/Sep/21

$$\underset{n\to \mathrm{\infty }}{\lim }.\frac{1}{n}.\frac{\underset{k=1}{\sum ^n}{\left(\frac{k}{n+1}\right)}^a}{n^{2}a+\frac{n(n+1)}{2}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\underset{k=1}{\sum ^n}{k}^a}{n{(n+1)}^a\left(\frac{2n^{2}a+n^2+n}{2}\right)}=0,\mathrm{\forall }a\in \mathbb{N}$$$$\underset{k=1}{\sum ^n}{k}^a=p\left(n\right),deg\left(p\right)=a+1,a\in \mathbb{N}$$$$$$

Commented by SANOGO last updated on 03/Sep/21

$$mercibienchef$$

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