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Question Number 32485 by abdo imad last updated on 25/Mar/18
let give α>1 find lim_(n→∞) Σ_(k=n+1) ^(2n) (1/k^α ) .
$${let}\:{give}\:\alpha>\mathrm{1}\:{find}\:{lim}_{{n}\rightarrow\infty} \:\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\:\frac{\mathrm{1}}{{k}^{\alpha} }\:. \\ $$
Commented by abdo imad last updated on 28/Mar/18
let put S_n = Σ_(k=n+1) ^(2n) (1/k^α ) S_n = (1/((n+1)^α )) +(1/((n+2)^α )) + ...+(1/((2n)^α )) =ξ_(2n) (α) −ξ_n (α) with ξ_n (x)=Σ_(k=1) ^n (1/k^x ) but lim_(n→∞) ξ_(2n) (α)=ξ(α) and lim_(n→∞) ξ_n (α)=ξ(α) ⇒ lim_(n→∞) S_n =0 anther method we have n+1 ≤ k≤ 2n ⇒ (1/(2n))≤ (1/k) ≤ (1/(n+1))≤ (1/n) ⇒ (1/((2n)^α )) ≤ (1/k^α ) ≤ (1/n^α ) ⇒ (n/((2n)^α )) ≤ Σ_(k=n+1) ^(2n) (1/k^α ) ≤ (n/n^α ) ⇒ (1/(2^α n^(α−1) )) ≤ S_n ≤ (1/n^(α−1) ) but α>1 ⇒ lim_(n→∞) (1/(2^α n^(α−1) )) =0 and lim_(n→∞) (1/n^(α−1) ) = 0 ⇒ lim_(n→∞) S_n =0
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\alpha} } \\ $$$${S}_{{n}} =\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\alpha} }\:\:+\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\alpha} }\:+\:…+\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\alpha} } \\ $$$$=\xi_{\mathrm{2}{n}} \left(\alpha\right)\:−\xi_{{n}} \left(\alpha\right)\:{with}\:\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:{but}\:{lim}_{{n}\rightarrow\infty} \xi_{\mathrm{2}{n}} \left(\alpha\right)=\xi\left(\alpha\right) \\ $$$${and}\:{lim}_{{n}\rightarrow\infty} \:\xi_{{n}} \left(\alpha\right)=\xi\left(\alpha\right)\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\mathrm{0} \\ $$$${anther}\:{method} \\ $$$${we}\:{have}\:\:\:\:{n}+\mathrm{1}\:\leqslant\:{k}\leqslant\:\mathrm{2}{n}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant\:\frac{\mathrm{1}}{{k}}\:\leqslant\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\leqslant\:\frac{\mathrm{1}}{{n}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\alpha} }\:\leqslant\:\:\frac{\mathrm{1}}{{k}^{\alpha} }\:\leqslant\:\:\frac{\mathrm{1}}{{n}^{\alpha} }\:\Rightarrow\:\:\frac{{n}}{\left(\mathrm{2}{n}\right)^{\alpha} }\:\leqslant\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\alpha} }\:\leqslant\:\:\frac{{n}}{{n}^{\alpha} }\:\Rightarrow \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\alpha} \:{n}^{\alpha−\mathrm{1}} }\:\leqslant\:\:{S}_{{n}} \:\leqslant\:\:\frac{\mathrm{1}}{{n}^{\alpha−\mathrm{1}} }\:\:{but}\:\alpha>\mathrm{1}\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{2}^{\alpha} \:{n}^{\alpha−\mathrm{1}} }\:=\mathrm{0}\:{and} \\ $$$${lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}^{\alpha−\mathrm{1}} }\:\:=\:\mathrm{0}\:\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\mathrm{0} \\ $$

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