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STATEMENT-1-For-every-natural-number-n-2-1-1-1-2-1-n-gt-n-and-STATEMENT-2-For-every-natural-number-n-2-n-n-1-lt-n-1-




Question Number 19215 by Tinkutara last updated on 07/Aug/17
STATEMENT-1 : For every natural  number n ≥ 2, (1/( (√1))) + (1/( (√2))) + ..... (1/( (√n))) > (√n)  and  STATEMENT-2 : For every natural  number n ≥ 2, (√(n(n + 1))) < n + 1
$$\mathrm{STATEMENT}-\mathrm{1}\::\:\mathrm{For}\:\mathrm{every}\:\mathrm{natural} \\ $$$$\mathrm{number}\:{n}\:\geqslant\:\mathrm{2},\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:…..\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:>\:\sqrt{{n}} \\ $$$$\boldsymbol{\mathrm{and}} \\ $$$$\mathrm{STATEMENT}-\mathrm{2}\::\:\mathrm{For}\:\mathrm{every}\:\mathrm{natural} \\ $$$$\mathrm{number}\:{n}\:\geqslant\:\mathrm{2},\:\sqrt{{n}\left({n}\:+\:\mathrm{1}\right)}\:<\:{n}\:+\:\mathrm{1} \\ $$
Answered by 433 last updated on 07/Aug/17
    1)  n>a (a=1,2,3,...,n−1)  (√n)>(√a)⇒(1/( (√n)))<(1/( (√a)))  (1/( (√1)))+(1/( (√2)))+(1/( (√3)))+...+(1/( (√n)))>(1/( (√n)))+(1/( (√n)))+(1/( (√n)))+...+(1/( (√n)))=(n/( (√n)))=(√n)    2)  n+1>n⇒^(n≥2) (n+1)(n+1)>n(n+1)⇒(√((n+1)(n+1)))>(√(n(n+1)))  (√(n(n+1)))<(√((n+1)(n+1)))<∣n+1∣=^(n≥2) n+1
$$ \\ $$$$ \\ $$$$\left.\mathrm{1}\right) \\ $$$$\mathrm{n}>\mathrm{a}\:\left(\mathrm{a}=\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{n}−\mathrm{1}\right) \\ $$$$\sqrt{\mathrm{n}}>\sqrt{\mathrm{a}}\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}<\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}>\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}=\frac{\mathrm{n}}{\:\sqrt{\mathrm{n}}}=\sqrt{\mathrm{n}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{n}+\mathrm{1}>\mathrm{n}\overset{\mathrm{n}\geqslant\mathrm{2}} {\Rightarrow}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)>\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\Rightarrow\sqrt{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}>\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\sqrt{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}<\sqrt{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}<\mid\mathrm{n}+\mathrm{1}\mid\overset{\mathrm{n}\geqslant\mathrm{2}} {=}\mathrm{n}+\mathrm{1} \\ $$
Commented by Tinkutara last updated on 07/Aug/17
But is 2 correct explanation of 1?
$$\mathrm{But}\:\mathrm{is}\:\mathrm{2}\:\mathrm{correct}\:\mathrm{explanation}\:\mathrm{of}\:\mathrm{1}? \\ $$

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