Question Number 205580 by universe last updated on 25/Mar/24
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$
Commented by lepuissantcedricjunior last updated on 26/Mar/24
![lim_(n→∞) (((2n+1)×(2n+3)×...×(4n+1))/((2n)×(2n+2)×..×(4n))) =lim_(n→∞) (((2n)(2n+1)(2n+2)...(4n)(4n+1))/([(2n)(2n+2)....(4n)]^2 )) =lim_(n→∞) ((Π_(k=0) ^(2n+1) (2n+k))/([2^n (n(n+1)(n+2)...(2n)]^2 )) =lim_(n→∞) ((n^(2n+2) Π_(k=0) ^(2n+1) (2+(k/n)))/(2^(2n) (Π_(p=0) ^n (n+k))^2 )) =lim_(n→∞) ((n^(2n+2) Π_(k=0) ^(2n+1) (2+(k/n)))/(2^(2n) (n^(n+1) Π_(k=0) ^n (1+(k/n)))^2 )) =lim_(n→∞) ((Π^(2n+1) (2+(k/n)))/(2^(2n) (Π_(k=0) ^n (1+(k/n)))^2 )) =lim_(n→∞) ((2+((2n+1)/n))/2^(2n+2) )≈lim_(n→∞) ((4+(1/n))/2^(2n+2) )=0](https://www.tinkutara.com/question/Q205658.png)
$$\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)×\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{3}\right)×…×\left(\mathrm{4}\boldsymbol{{n}}+\mathrm{1}\right)}{\left(\mathrm{2}\boldsymbol{{n}}\right)×\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)×..×\left(\mathrm{4}\boldsymbol{{n}}\right)} \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}\boldsymbol{{n}}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)…\left(\mathrm{4}\boldsymbol{{n}}\right)\left(\mathrm{4}\boldsymbol{{n}}+\mathrm{1}\right)}{\left[\left(\mathrm{2}\boldsymbol{{n}}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)….\left(\mathrm{4}\boldsymbol{{n}}\right)\right]^{\mathrm{2}} } \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}} {\prod}}\left(\mathrm{2}\boldsymbol{{n}}+\boldsymbol{{k}}\right)}{\left[\mathrm{2}^{\boldsymbol{{n}}} \left(\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(\boldsymbol{{n}}+\mathrm{2}\right)…\left(\mathrm{2}\boldsymbol{{n}}\right)\right]^{\mathrm{2}} \right.} \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\boldsymbol{{n}}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} \underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}} {\prod}}\left(\mathrm{2}+\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)}{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}} \left(\underset{\boldsymbol{{p}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\prod}}\left(\boldsymbol{{n}}+\boldsymbol{{k}}\right)\right)^{\mathrm{2}} } \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\boldsymbol{{n}}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} \underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}} {\prod}}\left(\mathrm{2}+\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)}{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}} \left(\boldsymbol{{n}}^{\boldsymbol{{n}}+\mathrm{1}} \underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\prod}}\left(\mathrm{1}+\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)\right)^{\mathrm{2}} } \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\overset{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}} {\prod}\left(\mathrm{2}+\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)}{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}} \left(\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\prod}}\left(\mathrm{1}+\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)\right)^{\mathrm{2}} } \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}+\frac{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}}{\boldsymbol{{n}}}}{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} }\approx\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}+\frac{\mathrm{1}}{\boldsymbol{{n}}}}{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} }=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 26/Mar/24
$${Please}\:{post}\:{your}\:{solution}\:{as}\: \\ $$$${answer}\:{instead}\:{as}\:{comment}\:{to} \\ $$$${the}\:{question}. \\ $$
Answered by Rasheed.Sindhi last updated on 28/Mar/24
$$\mathcal{T}{rying}…{not}\:{sure}… \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)…\left(\mathrm{2}{n}+\mathrm{2}{n}+\mathrm{1}\right)…\left({n}+\mathrm{1}\right)\:\mathrm{factors}}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)…\left(\mathrm{2}{n}+\mathrm{2}{n}\right)…\left({n}+\mathrm{1}\right)\:\mathrm{factors}}\:\: \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{2}}\centerdot\frac{\mathrm{2}{n}+\mathrm{5}}{\mathrm{2}{n}+\mathrm{4}}\centerdot…\centerdot\frac{\mathrm{2}{n}+\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}{n}}\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}\centerdot\frac{\mathrm{2}{n}+\mathrm{2}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\centerdot\frac{\mathrm{2}{n}+\mathrm{4}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{4}}\centerdot…\centerdot\frac{\mathrm{2}{n}+\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}{n}}\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{4}}\right)…\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}}\right)\right) \\ $$$$=\left(\mathrm{1}+\mathrm{0}\right)\left(\mathrm{1}+\mathrm{0}\right)\left(\mathrm{1}+\mathrm{0}\right)…\left(\mathrm{1}+\mathrm{0}\right) \\ $$$$=\mathrm{1} \\ $$