n-1-1-1-n-2-1-2-n-

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

Π_(n=1) ^∞ (( (1+ (1/n))^2 )/( 1+ (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

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

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