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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 byTinkutara 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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