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A-sequence-is-given-by-3-2-2-3-5-4-4-5-Write-down-the-general-term-of-the-sequence-and-find-its-limit-

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Question Number 4445 by Rasheed Soomro last updated on 28/Jan/16
A sequence is given by (3/2),(2/3),(5/4),(4/5),.... Write down the general term of the sequence and find its limit.
$$\mathrm{A}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},…. \\ $$$$\mathrm{Write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sequence}\:\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{limit}. \\ $$
Commented by 123456 last updated on 28/Jan/16
c_n =(a_n /b_n ) a_n = { ((3,n=1)),((a_(n−1) −1,n>1,n≡0(mod2))),((a_(n−1) +3,n>1,n≡1(mod2))) :} b_n =n+1 (3/2),(2/3),(5/4),(4/5),(7/6),(6/7),(9/8),(8/9),((11)/(10)),((10)/(11)),...
$${c}_{{n}} =\frac{{a}_{{n}} }{{b}_{{n}} } \\ $$$${a}_{{n}} =\begin{cases}{\mathrm{3},{n}=\mathrm{1}}\\{{a}_{{n}−\mathrm{1}} −\mathrm{1},{n}>\mathrm{1},{n}\equiv\mathrm{0}\left(\mathrm{mod2}\right)}\\{{a}_{{n}−\mathrm{1}} +\mathrm{3},{n}>\mathrm{1},{n}\equiv\mathrm{1}\left(\mathrm{mod2}\right)}\end{cases} \\ $$$${b}_{{n}} ={n}+\mathrm{1} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{7}}{\mathrm{6}},\frac{\mathrm{6}}{\mathrm{7}},\frac{\mathrm{9}}{\mathrm{8}},\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{11}}{\mathrm{10}},\frac{\mathrm{10}}{\mathrm{11}},… \\ $$
Commented by prakash jain last updated on 29/Jan/16
n^(th) term=((n+1+(−1)^(n+1) )/(n+1)) a_1 =((1+1+1)/2)=(3/2) a_2 =((2+1−1)/3)=(2/3) a_3 =((3+1+1)/4)=(5/4) a_4 =((4+1−1)/5)=(4/5)
$${n}^{{th}} \:\mathrm{term}=\frac{{n}+\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{1}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{2}+\mathrm{1}−\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{3}+\mathrm{1}+\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{4}+\mathrm{1}−\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$
Answered by prakash jain last updated on 29/Jan/16
a_n =((n+1+(−1)^(n+1) )/(n+1))=1+(((−1)^(n+1) )/(n+1)) lim_(n→∞) a_n =1
$${a}_{{n}} =\frac{{n}+\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\mathrm{1} \\ $$

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