STATEMENT - 1 : For every natural number [n 2,] [ 1 1 + 1 2 + ........

Question

STATEMENT - 1 : For every natural number $$n \ge 2,$$
$$ ${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + ........ + {1 \over {\sqrt n }} > \sqrt n .$$ $
STATEMENT - 2 : For every natural number  $$n \ge 2,$$,
\[

Accepted Answer

Statements $2$ is $$\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2$$ $$ \Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2$$ which is true $$ \Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n $$ Now $$\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}$$ $$\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};$$ $$\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}$$ Also $${1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}$$ $$\therefore$$ Adding all, we get $${1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n $$ Hence both the statements are correct and statement $2$ is a correct explanation of statement $-1.$

Solutions · 1