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Question-205772




Question Number 205772 by mr W last updated on 30/Mar/24
Answered by MM42 last updated on 30/Mar/24
S_n =((√(1×2))−1)  +((√(2×3))−(√(1×2))−1)  +((√(3×4))−(√(2×3))−1)  ⋮  +((√(n(n+1)))−(√((n−1)n))−1)  +((√((n+1)(n+2)))−(√(n(n+1)))−1)  ⇒S_n =(√(n^2 +3n+2))−n−1  ⇒lim_(n→∞)  S_n  =  (1/2)✓
$${S}_{{n}} =\left(\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{2}×\mathrm{3}}−\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{3}×\mathrm{4}}−\sqrt{\mathrm{2}×\mathrm{3}}−\mathrm{1}\right) \\ $$$$\vdots \\ $$$$+\left(\sqrt{{n}\left({n}+\mathrm{1}\right)}−\sqrt{\left({n}−\mathrm{1}\right){n}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\sqrt{{n}\left({n}+\mathrm{1}\right)}−\mathrm{1}\right) \\ $$$$\Rightarrow{S}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−{n}−\mathrm{1} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\checkmark \\ $$$$ \\ $$
Answered by A5T last updated on 30/Mar/24
T_0 =(√2)−0−1  T_(n−1) =(√(n^2 +n))−(√(n^2 −n))−1  T_n =(√(n^2 +3n+2))−(√(n^2 +n))−1  S_n =−(n+1)+(√(n^2 +3n+2))  n+1+n+2>2(√((n+1)(n+2)))⇒(√(n^2 +3n+2))<((2n+3)/2)  ⇒S_n =(√(n^2 +3n+2))−n−1<n+(3/2)−n−1=(1/2)  Question⇒lim_(n→∞) S_n =(1/2)
$${T}_{\mathrm{0}} =\sqrt{\mathrm{2}}−\mathrm{0}−\mathrm{1} \\ $$$${T}_{{n}−\mathrm{1}} =\sqrt{{n}^{\mathrm{2}} +{n}}−\sqrt{{n}^{\mathrm{2}} −{n}}−\mathrm{1} \\ $$$${T}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−\sqrt{{n}^{\mathrm{2}} +{n}}−\mathrm{1} \\ $$$${S}_{{n}} =−\left({n}+\mathrm{1}\right)+\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}} \\ $$$${n}+\mathrm{1}+{n}+\mathrm{2}>\mathrm{2}\sqrt{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\Rightarrow\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}<\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{S}_{{n}} =\sqrt{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}−{n}−\mathrm{1}<{n}+\frac{\mathrm{3}}{\mathrm{2}}−{n}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Question}\Rightarrow\underset{{n}\rightarrow\infty} {{lim}S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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