Question Number 4445 by Rasheed Soomro last updated on 28/Jan/16
$$\mathrm{A}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},…. \\ $$$$\mathrm{Write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sequence}\:\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{limit}. \\ $$
Commented by 123456 last updated on 28/Jan/16
$${c}_{{n}} =\frac{{a}_{{n}} }{{b}_{{n}} } \\ $$$${a}_{{n}} =\begin{cases}{\mathrm{3},{n}=\mathrm{1}}\\{{a}_{{n}−\mathrm{1}} −\mathrm{1},{n}>\mathrm{1},{n}\equiv\mathrm{0}\left(\mathrm{mod2}\right)}\\{{a}_{{n}−\mathrm{1}} +\mathrm{3},{n}>\mathrm{1},{n}\equiv\mathrm{1}\left(\mathrm{mod2}\right)}\end{cases} \\ $$$${b}_{{n}} ={n}+\mathrm{1} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{7}}{\mathrm{6}},\frac{\mathrm{6}}{\mathrm{7}},\frac{\mathrm{9}}{\mathrm{8}},\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{11}}{\mathrm{10}},\frac{\mathrm{10}}{\mathrm{11}},… \\ $$
Commented by prakash jain last updated on 29/Jan/16
$${n}^{{th}} \:\mathrm{term}=\frac{{n}+\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{1}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{2}+\mathrm{1}−\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{3}+\mathrm{1}+\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{4}+\mathrm{1}−\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$
Answered by prakash jain last updated on 29/Jan/16
$${a}_{{n}} =\frac{{n}+\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\mathrm{1} \\ $$