Question Number 31509 by abdo imad last updated on 09/Mar/18
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:+\mathrm{2}{kn}}}\:\:{find}\:\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} . \\ $$
Commented by abdo imad last updated on 13/Mar/18
![we have S_n =(1/n)Σ_(k=1) ^n (1/( (√(1+((2k)/n))))) =(1/2) (2/n)Σ_(k=1) ^n (1/( (√(1+((k(2−0))/n))))) ⇒lim_(n→∞) S_n =(1/2) ∫_0 ^(2 ) (dx/( (√(1+x))))=∫_0 ^2 (dx/(2(√(1+x))))=[(√(1+x)) ]_0 ^2 =(√3) −1 .](https://www.tinkutara.com/question/Q31756.png)
$${we}\:{have}\:{S}_{{n}} =\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{k}}{{n}}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}\left(\mathrm{2}−\mathrm{0}\right)}{{n}}}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\:\:\:} \:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}}}=\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{dx}}{\mathrm{2}\sqrt{\mathrm{1}+{x}}}=\left[\sqrt{\mathrm{1}+{x}}\:\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{3}}\:−\mathrm{1}\:. \\ $$