Menu Close

Name-Zainab-Bibi-BC200400692-Assignmeng-No-2-Mth-621-solution-a-n-n-n-2-a-n-n-n-2-n-n-2-2-by-ratio-test-a-n-a-n-n-n-2-

Tinku Tara 0 Comments
Pin




Question Number 194851 by Guriya last updated on 17/Jul/23
Name Zainab Bibi BC200400692 Assignmeng No#2 Mth 621 solution.. a_n =(((n− )!)/((n+ )^2 )) a_(n+ ) =(((n+ − )!)/((n+ + )^2 ))=(((n)!)/((n+2)^2 )) by ratio test (a_(n+ ) /a_n )=((((n)!)/((n+2)))/(((n− )!)/((n+ )^2 )))=(((n)!)/((n+2)^2 ))×(((n+ )^2 )/((n+ )!)) lim_(n→∞) (a_(n+ ) /a_n )=lim_(n→∞) ((n(n− )!)/((n+2)^2 ))×(((n+ )^2 )/((n− )!)) ∵(n)!=n(n− )! lim_(n→∞) (a_(n+ ) /a_n )=lim_(n→∞) ((n(n+ )^2 )/((n+2)^2 ))=lim_(n→∞) ((n(n^2 +1+2n))/(n^2 +4+4n)) lim_(n→∞) (((n^3 +n+2n^2 ))/(n^2 +4+4n))=lim_(n→∞) (((1+(1/n^2 )+(2/n)))/((1/n)+(1/n^3 )+(4/n^2 ))) Divided by n^3 to numerator and denominator Now by Applying limit (((1+(1/∞^(2 ) )+(2/∞)))/((1/∞)+(4/∞^3 )+(4/∞^2 )))=((1+0+0)/(0+0+0))=(1/0)=∞ lim_(n→∞) (a_(n+1) /a_n )=∞
$$\boldsymbol{\mathrm{Name}}\:\:\:\boldsymbol{\mathrm{Zainab}}\:\boldsymbol{\mathrm{Bibi}} \\ $$$$\boldsymbol{\mathrm{BC}}\mathrm{200400692} \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ssignmeng}\:\mathrm{No}#\mathrm{2}\: \\ $$$$\boldsymbol{\mathrm{Mth}}\:\mathrm{621} \\ $$$$\mathrm{solution}.. \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} } \\ $$$$\mathrm{a}_{\mathrm{n}+ } =\frac{\left(\mathrm{n}+ − \right)!}{\left(\mathrm{n}+ + \right)^{\mathrm{2}} }=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{by}\:\mathrm{ratio}\:\mathrm{test} \\ $$$$\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\frac{\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)}}{\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} }}=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+ \right)!} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\:\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}− \right)!}\:\:\:\:\:\because\left(\mathrm{n}\right)!=\mathrm{n}\left(\mathrm{n}− \right)! \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}+\mathrm{2n}\right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{n}^{\mathrm{3}} +\mathrm{n}+\mathrm{2n}^{\mathrm{2}} \right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}}=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{n}}\right)}{\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }} \\ $$$$\mathrm{Divided}\:\mathrm{by}\:\mathrm{n}^{\mathrm{3}} \:\mathrm{to}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator}\: \\ $$$$\mathrm{Now}\:\mathrm{by}\:\mathrm{Applying}\:\mathrm{limit} \\ $$$$\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\infty^{\mathrm{2}\:} }+\frac{\mathrm{2}}{\infty}\right)}{\frac{\mathrm{1}}{\infty}+\frac{\mathrm{4}}{\infty^{\mathrm{3}} }+\frac{\mathrm{4}}{\infty^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{0}+\mathrm{0}}{\mathrm{0}+\mathrm{0}+\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }=\infty \\ $$
Commented by Frix last updated on 17/Jul/23
(1/0)=∞ is not allowed; instead use this: (a_(n+ ) /a_n )=((((n)!)/((n+2)))/(((n− )!)/((n+ )^2 )))=(((n)!)/((n+2)^2 ))×(((n+ )^2 )/((n− )!))= =n×((n^2 +2n+1)/(n^2 +4n+4))=n×(1+((2n+1)/(n^2 +4n+1))) lim_(n→∞) n =∞ lim_(n→∞) 1 =1 lim_(n→∞) ((2n+1)/(n^2 +4n+1)) =0 ∞×(1+0)=∞
$$\frac{\mathrm{1}}{\mathrm{0}}=\infty\:\mathrm{is}\:\mathrm{not}\:\mathrm{allowed};\:\mathrm{instead}\:\mathrm{use}\:\mathrm{this}: \\ $$$$\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\frac{\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)}}{\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} }}=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}− \right)!}= \\ $$$$={n}×\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{4}}={n}×\left(\mathrm{1}+\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{1}}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\:=\infty \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}\:=\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{1}}\:=\mathrm{0} \\ $$$$\infty×\left(\mathrm{1}+\mathrm{0}\right)=\infty \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *