proof-that-2-n-1-lt-1-1-2-1-3-1-n-lt-2-n-

Question Number 36218 by Rio Mike last updated on 30/May/18

proof that

2 (\sqrt{n} - 1) < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots +

\frac{1}{\sqrt{n}} < 2 \sqrt{n}

Commented by abdo mathsup 649 cc last updated on 30/May/18

t h e f u n c t i o n f (x) = \frac{1}{\sqrt{x}} i s d e c r e a s i n g o n] 0, + \infty [s o

\int_{k}^{k + 1} f (t) d t ⩽ f (k) ⩽ \int_{k - 1}^{k} f (t) d t \Rightarrow

\sum_{k = 1}^{n} \int_{k}^{k + 1} f (t) d t ⩽ \sum_{k = 1}^{n} f (k) ⩽ \sum_{k = 1}^{n} \int_{n - 1}^{n} f (t) d t \Rightarrow

\int_{1}^{n + 1} \frac{d t}{\sqrt{t}} ⩽ 1 + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} ⩽ \int_{0}^{n} \frac{d t}{\sqrt{t}} \Rightarrow

{[2 \sqrt{t}]}_{1}^{n + 1} ⩽ 1 + \frac{1}{\sqrt{2}} + \dots . + \frac{1}{\sqrt{n}} - ⩽ {[2 . \sqrt{n}]}_{0}^{n} \Rightarrow

2 {\sqrt{n + 1} - 1} ⩽ 1 + \frac{1}{2} + \dots + \frac{1}{\sqrt{n}} ⩽ 2 \sqrt{n} b u t

\sqrt{n_{}} ⩽ . \sqrt{n + 1} \Rightarrow

2 {\sqrt{n} - 1} ⩽ 1 + \frac{1}{2} + \dots . + \frac{1}{\sqrt{n}} ⩽ 2 \sqrt{n} .