Prove Inequality: $\sqrt{1+\sqrt{2+...+\sqrt{2006}}} < 2$

[lfdahl]

Prove, that$\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{2006}}}}<2.$

 

[kaliprasad]

Prove, that$\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{2006}}}}<2.$

My Solution:

Spoiler

we have
$n = \sqrt{n^2} = \sqrt{n+n^2-n}\cdots(1)$
for $n>2$ $n^2-n > n > \sqrt{n+1}$ as $n^2 > 2n > n+1$
from above 2 we get
$n > \sqrt{n+\sqrt{n+1}}\cdots(1)$ for n > 2
for n = 2 we have $2 = \sqrt{2 + 2} = \sqrt{2 +\sqrt{3}}$ so for 2 also (1) holds

we have $2= \sqrt{1+3} > \sqrt{1 + 2} > \sqrt{1 + \sqrt{2 + \sqrt{3}}}$
applying (1) repeatedly until n = 2005 we get the result.

 

[lfdahl]

My Solution:

Spoiler

we have
$n = \sqrt{n^2} = \sqrt{n+n^2-n}\cdots(1)$
for $n>2$ $n^2-n > n > \sqrt{n+1}$ as $n^2 > 2n > n+1$
from above 2 we get
$n > \sqrt{n+\sqrt{n+1}}\cdots(1)$ for n > 2
for n = 2 we have $2 = \sqrt{2 + 2} = \sqrt{2 +\sqrt{3}}$ so for 2 also (1) holds

we have $2= \sqrt{1+3} > \sqrt{1 + 2} > \sqrt{1 + \sqrt{2 + \sqrt{3}}}$
applying (1) repeatedly until n = 2005 we get the result.

What a nice solution! Thankyou for your participation, kaliprasad!