Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 154441 by talminator2856791 last updated on 18/Sep/21

Π_(n=1) ^∞ (( (1+ (1/n))^2 )/( 1+ (2/n) ))

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\:\left(\mathrm{1}+\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} \:}{\:\mathrm{1}+\:\frac{\mathrm{2}}{{n}}\:} \\ $$$$\: \\ $$

Commented by benhamimed last updated on 18/Sep/21

=Π(((n+1)/n))^2 ×(n/(n+2))=Π(((n+1)^2 )/(n(n+2))) =lim((2^2 ×3^2 ×4^2 ×..×(n−1)^2 ×n^2 ×(n+1)^2 )/((1×3)(2×4)(3×5)×..×((n−2)n)((n−1)(n+1))(n(n+2)))) lim((2(n+1))/((n+2)))=2

$$=\Pi\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{\mathrm{2}} ×\frac{{n}}{{n}+\mathrm{2}}=\Pi\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{{n}\left({n}+\mathrm{2}\right)} \\ $$$$={lim}\frac{\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{4}^{\mathrm{2}} ×..×\left({n}−\mathrm{1}\right)^{\mathrm{2}} ×{n}^{\mathrm{2}} ×\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}×\mathrm{3}\right)\left(\mathrm{2}×\mathrm{4}\right)\left(\mathrm{3}×\mathrm{5}\right)×..×\left(\left({n}−\mathrm{2}\right){n}\right)\left(\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)\right)\left({n}\left({n}+\mathrm{2}\right)\right)} \\ $$$${lim}\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{2}\right)}=\mathrm{2} \\ $$

Answered by Kamel last updated on 18/Sep/21

Π_(n=1) ^∞ (( (1+ (1/n))^2 )/( 1+ (2/n) )) P=Π_(n=1) ^(+∞) ((n+1)/n).((n+1)/(n+2))=(2/1).(2/3).(3/2).(3/4).(4/3).(4/5)... =2

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\:\left(\mathrm{1}+\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} \:}{\:\mathrm{1}+\:\frac{\mathrm{2}}{{n}}\:} \\ $$$$\:{P}=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\prod}}\frac{{n}+\mathrm{1}}{{n}}.\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{1}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}.\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{4}}{\mathrm{5}}... \\ $$$$\:\:\:\:=\mathrm{2} \\ $$

Commented by mnjuly1970 last updated on 18/Sep/21

ln (p )= lim Σ_(k=1) ^n {ln(((k+1)/k))+ln(((k+1)/(k+2)))} = lim_(n→∞) (n +1 ) +ln (2) −lim_(n→∞) (n+2) = lim_(n→∞) (((n+1)/(n+2))) +ln(2)=ln(2) ∴ p=2

$$\:\:{ln}\:\left({p}\:\right)=\:{lim}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{{ln}\left(\frac{{k}+\mathrm{1}}{{k}}\right)+{ln}\left(\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}}\right)\right\} \\ $$$$\:\:=\:{lim}_{{n}\rightarrow\infty} \:\left({n}\:+\mathrm{1}\:\right)\:+{ln}\:\left(\mathrm{2}\right)\:−{lim}_{{n}\rightarrow\infty} \left({n}+\mathrm{2}\right) \\ $$$$\:\:\:=\:{lim}_{{n}\rightarrow\infty} \left(\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\right)\:+{ln}\left(\mathrm{2}\right)={ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:{p}=\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com