Previous in Relation and Functions Next in Relation and Functions
Question Number 6238 by sanusihammed last updated on 19/Jun/16
Answered by Yozzii last updated on 20/Jun/16
![Let lim_(n→∞) (1+n^(−1) )^(−n^2 ) =l. Let u=n^(−1) ⇒l=lim_(u→0^+ ) (1+u)^(−(1/u^2 )) ⇒lnl=lim_(u→0^+ ) {((−1)/u^2 )ln(1+u)} lnl=lim_(u→0^+ ) (((−1)/u^2 )(u−(u^2 /2)+(u^3 /3)−(u^4 /4)+(u^5 /5)−...)) lnl=lim_(u→0^+ ) (((−1)/u)+(1/2)−(u/3)+(u^2 /4)−(u^3 /5)+...) lnl=−∞⇒l=e^(−∞) =0. ∴Σ(1+n^(−1) )^(−n^2 ) is possibly convergent. Ratio test: let u(n)=(1+n^(−1) )^(−n^2 ) . ∴ r=lim_(n→∞) ((u(n+1))/(u(n)))=lim_(n→∞) (((1+(1/(n+1)))^(−(n+1)^2 ) )/((1+(1/n))^(−n^2 ) )) Let u=(1/n)⇒n+1=((1+u)/u)⇒(1/(n+1))=(u/(u+1)) r=lim_(u→0^+ ) (((1+(u/(u+1)))^(−(((u+1)/u))^2 ) )/((1+u)^(−(1/u^2 )) )) r=lim_(u→0^+ ) {(((1+(u/(u+1)))^((u+1)^2 ) )/(1+u))}^(−1/u^2 ) r=lim_(u→0^+ ) {(((2u+1)^((u+1)^2 ) )/((u+1)^((u+1)^2 +1) ))}^(−1/u^2 ) ⇒lnr=lim_(u→0^+ ) {((−1)/u^2 )[(u+1)^2 ln(1+2u)−(u^2 +2u+2)ln(1+u)]} lnr=lim_(u→0^+ ) {(1+(2/u)+(2/u^2 ))ln(1+u)−(1+(2/u)+(1/u^2 ))ln(1+2u)} lnr=lim_(u→0^+ ) {ln((1+u)/(1+2u))+(2/u)(ln(1+u)−ln(1+2u))+(1/u^2 )(2ln(1+u)−ln(1+2u))} −−−−−−−−−−−−−−−−−−−−−−−−−−− ln(1+u)−ln(1+2u)=u−(u^2 /2)+(u^3 /3)−(u^4 /4)+(u^5 /5)−(u^6 /6)+...−(2u−((4u^2 )/2)+((8u^3 )/3)−((16u^4 )/4)+((32u^5 )/5)−((64u^6 )/6)+...) ln(1+u)−ln(1+2u)=(1−2)u+(u^2 /2)(−1+4)+(u^3 /3)(1−8)+(u^4 /4)(16−1)+(u^5 /5)(−32+1)+(u^6 /6)(64−1)+... ln((1+u)/(1+2u))=Σ_(i=1) ^∞ (−1)^i (u^i /i)(2^i −1) ⇒lim_(u→0^+ ) ((2/u)ln((1+u)/(1+2u)))=lim_(u→0^+ ) (2Σ_(i=1) ^∞ (−1)^i (2^i −1)(u^(i−1) /i) )=2(−1)(2−1)(1/1)=−2 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 2ln(1+u)−ln(1+2u)=2Σ_(i=1) ^∞ ((u^i (−1)^(i+1) )/i)−Σ_(i=1) ^∞ (((2u)^i (−1)^(i+1) )/i) ln(((1+u)^2 )/(1+2u))=2u−2u+Σ_(i=2) ^∞ (((−1)^(i+1) )/i)(2u^i −2^i u^i ) lim_(u→0^+ ) (1/u^2 )ln(((1+u)^2 )/(1+2u))=lim_(u→0^+ ) [(1/u^2 ){(−1)^3 (1/2)(2u^2 −2^2 u^2 )+Σ_(i=3) ^∞ ((((−1)^(i+1) (2u^i −2^i u^i ))/i))}] lim_(u→0^+ ) ((1/u^2 )ln(((1+u)^2 )/(1+2u)))=1 −−−−−−−−−−−−−−−−−−−−−−−−−− lnr=ln((1+0)/(1+0))−2+1=−1⇒r=e^(−1) or 0<r<1⇒ the series is convergent.](Q6244.png)
$${Let}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+{n}^{−\mathrm{1}} \right)^{−{n}^{\mathrm{2}} } ={l}. \\ $$$${Let}\:{u}={n}^{−\mathrm{1}} \Rightarrow{l}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{1}+{u}\right)^{−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\Rightarrow{lnl}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{\frac{−\mathrm{1}}{{u}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{u}\right)\right\} \\ $$$${lnl}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{−\mathrm{1}}{{u}^{\mathrm{2}} }\left({u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−...\right)\right) \\ $$$${lnl}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{−\mathrm{1}}{{u}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{u}}{\mathrm{3}}+\frac{{u}^{\mathrm{2}} }{\mathrm{4}}−\frac{{u}^{\mathrm{3}} }{\mathrm{5}}+...\right) \\ $$$${lnl}=−\infty\Rightarrow{l}={e}^{−\infty} =\mathrm{0}.\:\therefore\Sigma\left(\mathrm{1}+{n}^{−\mathrm{1}} \right)^{−{n}^{\mathrm{2}} } \\ $$$${is}\:{possibly}\:{convergent}. \\ $$$${Ratio}\:{test}:\:{let}\:{u}\left({n}\right)=\left(\mathrm{1}+{n}^{−\mathrm{1}} \right)^{−{n}^{\mathrm{2}} } . \\ $$$$\therefore\:{r}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{u}\left({n}+\mathrm{1}\right)}{{u}\left({n}\right)}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{−\left({n}+\mathrm{1}\right)^{\mathrm{2}} } }{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{−{n}^{\mathrm{2}} } } \\ $$$${Let}\:{u}=\frac{\mathrm{1}}{{n}}\Rightarrow{n}+\mathrm{1}=\frac{\mathrm{1}+{u}}{{u}}\Rightarrow\frac{\mathrm{1}}{{n}+\mathrm{1}}=\frac{{u}}{{u}+\mathrm{1}} \\ $$$${r}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{u}}{{u}+\mathrm{1}}\right)^{−\left(\frac{{u}+\mathrm{1}}{{u}}\right)^{\mathrm{2}} } }{\left(\mathrm{1}+{u}\right)^{−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} } \\ $$$${r}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{\frac{\left(\mathrm{1}+\frac{{u}}{{u}+\mathrm{1}}\right)^{\left({u}+\mathrm{1}\right)^{\mathrm{2}} } }{\mathrm{1}+{u}}\right\}^{−\mathrm{1}/{u}^{\mathrm{2}} } \\ $$$${r}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{\frac{\left(\mathrm{2}{u}+\mathrm{1}\right)^{\left({u}+\mathrm{1}\right)^{\mathrm{2}} } }{\left({u}+\mathrm{1}\right)^{\left({u}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} }\right\}^{−\mathrm{1}/{u}^{\mathrm{2}} } \\ $$$$\Rightarrow{lnr}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{\frac{−\mathrm{1}}{{u}^{\mathrm{2}} }\left[\left({u}+\mathrm{1}\right)^{\mathrm{2}} {ln}\left(\mathrm{1}+\mathrm{2}{u}\right)−\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2}\right){ln}\left(\mathrm{1}+{u}\right)\right]\right\} \\ $$$${lnr}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{\left(\mathrm{1}+\frac{\mathrm{2}}{{u}}+\frac{\mathrm{2}}{{u}^{\mathrm{2}} }\right){ln}\left(\mathrm{1}+{u}\right)−\left(\mathrm{1}+\frac{\mathrm{2}}{{u}}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){ln}\left(\mathrm{1}+\mathrm{2}{u}\right)\right\} \\ $$$${lnr}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left\{{ln}\frac{\mathrm{1}+{u}}{\mathrm{1}+\mathrm{2}{u}}+\frac{\mathrm{2}}{{u}}\left({ln}\left(\mathrm{1}+{u}\right)−{ln}\left(\mathrm{1}+\mathrm{2}{u}\right)\right)+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\left(\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)−{ln}\left(\mathrm{1}+\mathrm{2}{u}\right)\right)\right\} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${ln}\left(\mathrm{1}+{u}\right)−{ln}\left(\mathrm{1}+\mathrm{2}{u}\right)={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−\frac{{u}^{\mathrm{6}} }{\mathrm{6}}+...−\left(\mathrm{2}{u}−\frac{\mathrm{4}{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{8}{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{16}{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{32}{u}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{64}{u}^{\mathrm{6}} }{\mathrm{6}}+...\right) \\ $$$${ln}\left(\mathrm{1}+{u}\right)−{ln}\left(\mathrm{1}+\mathrm{2}{u}\right)=\left(\mathrm{1}−\mathrm{2}\right){u}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{1}+\mathrm{4}\right)+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\left(\mathrm{1}−\mathrm{8}\right)+\frac{{u}^{\mathrm{4}} }{\mathrm{4}}\left(\mathrm{16}−\mathrm{1}\right)+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\left(−\mathrm{32}+\mathrm{1}\right)+\frac{{u}^{\mathrm{6}} }{\mathrm{6}}\left(\mathrm{64}−\mathrm{1}\right)+... \\ $$$${ln}\frac{\mathrm{1}+{u}}{\mathrm{1}+\mathrm{2}{u}}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{{u}^{{i}} }{{i}}\left(\mathrm{2}^{{i}} −\mathrm{1}\right) \\ $$$$\Rightarrow\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{2}}{{u}}{ln}\frac{\mathrm{1}+{u}}{\mathrm{1}+\mathrm{2}{u}}\right)=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{2}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \left(\mathrm{2}^{{i}} −\mathrm{1}\right)\frac{{u}^{{i}−\mathrm{1}} }{{i}}\:\right)=\mathrm{2}\left(−\mathrm{1}\right)\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{1}/\mathrm{1}\right)=−\mathrm{2} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)−{ln}\left(\mathrm{1}+\mathrm{2}{u}\right)=\mathrm{2}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{u}^{{i}} \left(−\mathrm{1}\right)^{{i}+\mathrm{1}} }{{i}}−\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{u}\right)^{{i}} \left(−\mathrm{1}\right)^{{i}+\mathrm{1}} }{{i}} \\ $$$${ln}\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{u}}=\mathrm{2}{u}−\mathrm{2}{u}+\underset{{i}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} }{{i}}\left(\mathrm{2}{u}^{{i}} −\mathrm{2}^{{i}} {u}^{{i}} \right) \\ $$$$\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{ln}\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{u}}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\left\{\left(−\mathrm{1}\right)^{\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} {u}^{\mathrm{2}} \right)+\underset{{i}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} \left(\mathrm{2}{u}^{{i}} −\mathrm{2}^{{i}} {u}^{{i}} \right)}{{i}}\right)\right\}\right] \\ $$$$\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{ln}\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{u}}\right)=\mathrm{1} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${lnr}={ln}\frac{\mathrm{1}+\mathrm{0}}{\mathrm{1}+\mathrm{0}}−\mathrm{2}+\mathrm{1}=−\mathrm{1}\Rightarrow{r}={e}^{−\mathrm{1}} \\ $$$${or}\:\mathrm{0}<{r}<\mathrm{1}\Rightarrow\:{the}\:{series}\:{is}\:{convergent}. \\ $$$$ \\ $$$$ \\ $$
Commented by sanusihammed last updated on 20/Jun/16
$${Wow}\:{thanks}\:{so}\:{much} \\ $$