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Question Number 32485 by abdo imad last updated on 25/Mar/18

$$letgive\alpha >1find{lim}_{n\to \mathrm{\infty }}\sum _{k=n+1}^{2n}\frac{1}{k^{\alpha }}.$$

Commented byabdo imad last updated on 28/Mar/18

$$letput{S}_n=\sum _{k=n+1}^{2n}\frac{1}{k^{\alpha }}$$ $$S_n=\frac{1}{{(n+1)}^{\alpha }}+\frac{1}{{(n+2)}^{\alpha }}+...+\frac{1}{{\left(2n\right)}^{\alpha }}$$ $$={\xi }_{2n}\left(\alpha \right)-{\xi }_n\left(\alpha \right)with{\xi }_n\left(x\right)=\sum _{k=1}^n\frac{1}{k^x}but{lim}_{n\to \mathrm{\infty }}{\xi }_{2n}\left(\alpha \right)=\xi \left(\alpha \right)$$ $$and{lim}_{n\to \mathrm{\infty }}{\xi }_n\left(\alpha \right)=\xi \left(\alpha \right)\Rightarrow {lim}_{n\to \mathrm{\infty }}{S}_n=0$$ $$anthermethod$$ $$wehaven+1⩽k⩽2n\Rightarrow \frac{1}{2n}⩽\frac{1}{k}⩽\frac{1}{n+1}⩽\frac{1}{n}\Rightarrow$$ $$\frac{1}{{\left(2n\right)}^{\alpha }}⩽\frac{1}{k^{\alpha }}⩽\frac{1}{n^{\alpha }}\Rightarrow \frac{n}{{\left(2n\right)}^{\alpha }}⩽\sum _{k=n+1}^{2n}\frac{1}{k^{\alpha }}⩽\frac{n}{n^{\alpha }}\Rightarrow$$ $$\frac{1}{2^{\alpha }{n}^{\alpha -1}}⩽S_n⩽\frac{1}{n^{\alpha -1}}but\alpha >1\Rightarrow {lim}_{n\to \mathrm{\infty }}\frac{1}{2^{\alpha }{n}^{\alpha -1}}=0and$$ $${lim}_{n\to \mathrm{\infty }}\frac{1}{n^{\alpha -1}}=0\Rightarrow {lim}_{n\to \mathrm{\infty }}{S}_n=0$$

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