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let-S-n-k-1-2n-1-1-n-2-k-calculste-lim-n-S-n-

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Question Number 57409 by Abdo msup. last updated on 03/Apr/19
let S_n =Σ_(k=1) ^(2n+1) (1/( (√(n^2 +k)))) calculste lim_(n→+∞) S_n
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:+{k}}} \\ $$$${calculste}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 10/Apr/19
we have (1/( (√(n^2 +k)))) =(1/(n(√(1+(k/n^2 ))))) =(((1+(k/n^2 ))^(−(1/2)) )/n) but we have (1+u)^α =1+α u +((α(α−1))/2) u^2 +0(u^3 ) ⇒ (1+α)^(−(1/2)) =1−(α/2) +(((−(1/2))(−(1/2)−1))/2) α^(2 ) +o(α^3 ) =1−(α/2) +(3/8) α^2 +0(α^3 ) ⇒1−(α/2) ≤ (1+α)^(−(1/2)) ≤1−(α/2) +(3/8) α^2 ⇒ 1−(k/(2n^2 )) ≤ (1+(k/n^2 ))^(−(1/2)) ≤1−(k/(2n^2 )) +(3/8) (k^2 /n^4 ) ⇒(1/n) −(k/(2n^3 )) ≤(1/n)(1+(k/n^2 ))^(−(1/2)) ≤ (1/n) −(k/(2n^3 )) +(3/8) (k^2 /n^5 ) ⇒Σ_(k=1) ^(2n+1) ((1/n) −(k/(2n^3 ))) ≤ S_n ≤Σ_(k=1) ^(2n+1) ((1/n) −(k/(2n^3 )) +(3/8) (k^2 /n^5 )) ⇒ 2−(1/(2n^3 )) Σ_(k=1) ^(2n+1) k ≤ S_n ≤ 2−(1/(2n^3 )) Σ_(k=1) ^(2n+1) k +(3/(8n^5 )) Σ_(k=1) ^(2n+1) k^2 2−(1/(2n^3 )) (((2n+1)(2n+2))/2) ≤ S_n ≤ 2−(1/(2n^3 )) (((2n+1)(2n+2))/2) +(3/(8n^5 )) (((2n+1)(2n+2)(4n+2+1))/6) ⇒ 2−(((n+1)(2n+1))/(2n^3 )) ≤ S_n ≤ 2−(((n+1)(2n+1))/(2n^3 )) +(1/(8n^5 )) (n+1)(2n+1)(4n+3) ⇒ lim_(n→+∞) S_n =2 .
$${we}\:{have}\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:+{k}}}\:=\frac{\mathrm{1}}{{n}\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }}}\:=\frac{\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{{n}} \\ $$$${but}\:{we}\:{have}\:\left(\mathrm{1}+{u}\right)^{\alpha} \:=\mathrm{1}+\alpha\:{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}}\:{u}^{\mathrm{2}} \:+\mathrm{0}\left({u}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$$\left(\mathrm{1}+\alpha\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\mathrm{1}−\frac{\alpha}{\mathrm{2}}\:+\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}\:\alpha^{\mathrm{2}\:} \:+{o}\left(\alpha^{\mathrm{3}} \right) \\ $$$$=\mathrm{1}−\frac{\alpha}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\alpha^{\mathrm{2}} \:+\mathrm{0}\left(\alpha^{\mathrm{3}} \right)\:\Rightarrow\mathrm{1}−\frac{\alpha}{\mathrm{2}}\:\leqslant\:\left(\mathrm{1}+\alpha\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\leqslant\mathrm{1}−\frac{\alpha}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\alpha^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }\:\leqslant\:\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \leqslant\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{8}}\:\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{4}} }\:\Rightarrow\frac{\mathrm{1}}{{n}}\:−\frac{{k}}{\mathrm{2}{n}^{\mathrm{3}} }\:\leqslant\frac{\mathrm{1}}{{n}}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \leqslant \\ $$$$\frac{\mathrm{1}}{{n}}\:−\frac{{k}}{\mathrm{2}{n}^{\mathrm{3}} }\:+\frac{\mathrm{3}}{\mathrm{8}}\:\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{5}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\:−\frac{{k}}{\mathrm{2}{n}^{\mathrm{3}} }\right)\:\leqslant\:{S}_{{n}} \leqslant\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\:−\frac{{k}}{\mathrm{2}{n}^{\mathrm{3}} }\:+\frac{\mathrm{3}}{\mathrm{8}}\:\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{5}} }\right)\:\Rightarrow \\ $$$$\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }\:\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \:{k}\:\leqslant\:{S}_{{n}} \leqslant\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }\:\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} {k}\:\:+\frac{\mathrm{3}}{\mathrm{8}{n}^{\mathrm{5}} }\:\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \:{k}^{\mathrm{2}} \\ $$$$\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}}\:\leqslant\:{S}_{{n}} \leqslant\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}{n}^{\mathrm{5}} }\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{4}{n}+\mathrm{2}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{2}−\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{3}} }\:\leqslant\:{S}_{{n}} \leqslant\:\mathrm{2}−\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{5}} }\:\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right) \\ $$$$\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\mathrm{2}\:. \\ $$$$ \\ $$

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