Question Number 212544 by MrGaster last updated on 17/Oct/24
![lim_(n→∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ]=?](https://www.tinkutara.com/question/Q212544.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}^{\mathrm{2}} \left[\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right]=? \\ $$
Answered by lepuissantcedricjunior last updated on 17/Oct/24
![lim_(n→+∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ] =lim_(n→+∞) {n^2 (1+(1/(n+1)))^(n+1) −n^2 (1+(1/n))^n } posons : { (((1/(n+1))=a)),(((1/n)=b)) :}=> { ((n=(1/a)−1)),((n=(1/b))) :}=> { ((n→+∞;a→0)),((n→+∞;b→0)) :} =lim_((a;b)→(0;0)) {((1/a)−1)^2 e^((ln(1+a))/a) −(1/b^2 )e^((ln(1+b))/b) } =lim_((a;b)→(0;0)) {e^(((ln(1+a))/a)+2ln(1−a)−2lna) −e^(((ln(1+b))/b)+2lnb) } =lim_((a;b)→(0;0)) (1−(2/a))e^((ln(1+a))/a) =−∞×1=−∞ =>lim_(n→+∞) n^2 [(1+(1/(n+1)))^(n+1) −(1+(1/n))^n ]=−∞](https://www.tinkutara.com/question/Q212551.png)
$$\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{n}}^{\mathrm{2}} \left[\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}\right)^{\boldsymbol{{n}}+\mathrm{1}} −\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}}\right)^{\boldsymbol{{n}}} \right] \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\left\{\boldsymbol{{n}}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}\right)^{\boldsymbol{{n}}+\mathrm{1}} −\boldsymbol{{n}}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}}\right)^{\boldsymbol{{n}}} \right\} \\ $$$$\boldsymbol{{posons}}\::\begin{cases}{\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}=\boldsymbol{{a}}}\\{\frac{\mathrm{1}}{\boldsymbol{{n}}}=\boldsymbol{{b}}}\end{cases}=>\begin{cases}{\boldsymbol{{n}}=\frac{\mathrm{1}}{\boldsymbol{{a}}}−\mathrm{1}}\\{\boldsymbol{{n}}=\frac{\mathrm{1}}{\boldsymbol{{b}}}}\end{cases}=>\begin{cases}{\boldsymbol{{n}}\rightarrow+\infty;\boldsymbol{{a}}\rightarrow\mathrm{0}}\\{\boldsymbol{{n}}\rightarrow+\infty;\boldsymbol{{b}}\rightarrow\mathrm{0}}\end{cases} \\ $$$$=\underset{\left(\boldsymbol{{a}};\boldsymbol{{b}}\right)\rightarrow\left(\mathrm{0};\mathrm{0}\right)} {\mathrm{lim}}\left\{\left(\frac{\mathrm{1}}{\boldsymbol{{a}}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{a}}\right)}{\boldsymbol{{a}}}} −\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{2}} }\boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{b}}\right)}{\boldsymbol{{b}}}} \right\} \\ $$$$=\underset{\left(\boldsymbol{{a}};\boldsymbol{{b}}\right)\rightarrow\left(\mathrm{0};\mathrm{0}\right)} {\mathrm{lim}}\left\{\boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{a}}\right)}{\boldsymbol{{a}}}+\mathrm{2}\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{a}}\right)−\mathrm{2}\boldsymbol{{lna}}} −\boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{b}}\right)}{\boldsymbol{{b}}}+\mathrm{2}\boldsymbol{{lnb}}} \right\} \\ $$$$=\underset{\left(\boldsymbol{{a}};\boldsymbol{{b}}\right)\rightarrow\left(\mathrm{0};\mathrm{0}\right)} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{2}}{\boldsymbol{{a}}}\right)\boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{a}}\right)}{\boldsymbol{{a}}}} =−\infty×\mathrm{1}=−\infty \\ $$$$=>\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{n}}^{\mathrm{2}} \left[\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}+\mathrm{1}}\right)^{\boldsymbol{{n}}+\mathrm{1}} −\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{n}}}\right)^{\boldsymbol{{n}}} \right]=−\infty \\ $$
Commented by Ghisom last updated on 17/Oct/24
$$\mathrm{wrong} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} >\mathrm{0}\forall{n}\in\mathbb{Z}^{+} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{limit}\:\mathrm{cannot}\:\mathrm{be}\:<\mathrm{0} \\ $$