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Question Number 50405 by Abdo msup. last updated on 16/Dec/18

let  V_n = (1/(2n+1)) +(1/(2n+3)) +...+(1/(4n−1))  determine lim_(n→+∞)  V_n

$${let}\:\:{V}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\:+...+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$$${determine}\:{lim}_{{n}\rightarrow+\infty} \:{V}_{{n}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

V_n =(1/(2n+1))+(1/(2n+3))+...+(1/(2n+2n−1))  V_n =((1/n)/(2+(1/n)))+((1/n)/(2+(3/n)))+...+((1/n)/(2+((2n−1)/n)))  lim_(n→∞) (1/n)Σ_(r=1) ^n   (1/(1+(((2r−1)/n))))=lim_(n→∞) (1/n)Σ_(r=1) ^n ((1/(1+2((r/n))−(1/n)))  ∫_0 ^1 (dx/(1+2x))=(1/2)∫_0 ^1 (dx/((1/2)+x))=(1/2)∣ln((1/2)+x)∣_0 ^1   =(1/2)[ln((3/2))−ln((1/2))]=(1/2)ln3    pls check is it correct...

$${V}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}{n}−\mathrm{1}} \\ $$$${V}_{{n}} =\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{2}+\frac{\mathrm{1}}{{n}}}+\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{2}+\frac{\mathrm{3}}{{n}}}+...+\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{2}+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}} \\ $$$${li}\underset{{n}\rightarrow\infty} {{m}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{2}{r}−\mathrm{1}}{{n}}\right)}={li}\underset{{n}\rightarrow\infty} {{m}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\left(\frac{{r}}{{n}}\right)−\frac{\mathrm{1}}{{n}}}\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\frac{\mathrm{1}}{\mathrm{2}}+{x}}=\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right]=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3} \\ $$$$ \\ $$$${pls}\:{check}\:{is}\:{it}\:{correct}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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