Will difference always works?

Calculus Level 3

2 1 0.4142 3 2 0.3178 4 3 0.2679 \begin{aligned} & \sqrt2 - \sqrt 1 \approx 0.4142\\& \sqrt3 - \sqrt 2 \approx 0.3178\\& \sqrt4 - \sqrt3\approx 0.2679 \\& \vdots\quad\vdots\quad\vdots\end{aligned}

True or false

There exist positive integer n n such that n + 1 n > 3 5 \sqrt{n+1}- \sqrt n > \dfrac{3}{5} .

false True

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1 solution

Chew-Seong Cheong
Feb 17, 2018

X = n + 1 n = ( n + 1 n ) ( n + 1 + n ) n + 1 + n = n + 1 n n + 1 + n = 1 n + 1 + n \begin{aligned} X & = \sqrt{n+1}-\sqrt n \\ & = \frac {(\sqrt{n+1}-\sqrt n)( \sqrt{n+1}+\sqrt n)} {\sqrt{n+1}+\sqrt n} \\ & = \frac {n+1-n} {\sqrt{n+1}+\sqrt n} \\ & = \frac 1{\sqrt{n+1}+\sqrt n} \end{aligned} .

X X is the largest when n = 1 n=1 . Therefore, X 1 2 + 1 0.4142 < 3 5 X \le \dfrac 1{\sqrt 2 + 1} \approx 0.4142 < \dfrac 35 . The statement is false .

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