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, \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots . . \frac{1}{\sqrt{n}} > \sqrt{n}

and

STATEMENT - 2 : For every natural

number n ⩾ 2, \sqrt{n (n + 1)} < n + 1

Answered by 433 last updated on 07/Aug/17

1)

n > a (a = 1, 2, 3, \dots, n - 1)

\sqrt{n} > \sqrt{a} \Rightarrow \frac{1}{\sqrt{n}} < \frac{1}{\sqrt{a}}

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} > \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \dots + \frac{1}{\sqrt{n}} = \frac{n}{\sqrt{n}} = \sqrt{n}

2)

n + 1 > n \overset{n ⩾ 2}{\Rightarrow} (n + 1) (n + 1) > n (n + 1) \Rightarrow \sqrt{(n + 1) (n + 1)} > \sqrt{n (n + 1)}

\sqrt{n (n + 1)} < \sqrt{(n + 1) (n + 1)} <∣ n + 1 ∣ \overset{n ⩾ 2}{=} n + 1

Commented by Tinkutara last updated on 07/Aug/17

But is 2 correct explanation of 1 ?