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

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In summary, the conversation discusses the inequality $\sqrt{1+\sqrt{2+...+\sqrt{2006}}} < 2$ and how it can be proven using mathematical induction. It also mentions a geometric interpretation where the nested square roots can be seen as layers of a pyramid. The significance of the number 2006 is explained as an arbitrary choice for a more interesting result, and the generalization of the inequality to other values and patterns is also discussed. This inequality may have practical applications in fields such as physics and engineering.
  • #1
lfdahl
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Prove, that$\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{2006}}}}<2.$
 
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  • #2
lfdahl said:
Prove, that$\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{2006}}}}<2.$

My Solution:

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.
 
  • #3
kaliprasad said:
My Solution:

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!
 

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

How do you prove the inequality $\sqrt{1+\sqrt{2+...+\sqrt{2006}}} < 2$?

There are a few different ways to prove this inequality. One approach is to use mathematical induction, where you first show that the inequality holds for the base case of n=1. Then, you assume that it holds for some arbitrary value n=k, and use that assumption to prove that it also holds for n=k+1. This shows that the inequality holds for all natural numbers n, including n=2006.

Can you provide a geometric interpretation of this inequality?

Yes, there is a geometric interpretation of this inequality. If you imagine the nested square roots as layers of a pyramid, the sum inside the square root can be seen as the number of layers. As the number of layers increases, the overall value of the pyramid also increases. However, since the value of the pyramid is always less than 2, the square root of the pyramid must also be less than 2.

What is the significance of the number 2006 in this inequality?

The number 2006 is simply the arbitrary value chosen for this specific inequality. This inequality can be proven for any natural number n, but using 2006 as the value gives a more interesting and complex result.

Can this inequality be generalized to other values or patterns?

Yes, this inequality can be generalized to other values and patterns. For example, instead of starting with 1 in the first square root, we can start with any other positive number, and the inequality will still hold. Additionally, this inequality can be extended to a larger number of square roots, as long as the pattern remains the same.

How can this inequality be applied in real-life situations?

This inequality may have applications in fields such as physics and engineering, where nested square roots can appear in equations. By understanding and proving this inequality, scientists and engineers can have a better understanding of the behavior of these equations and make more accurate predictions.

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