IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

The fundamental theorem of algebra has no place for $\infty$ as it is not a real number. And notice I used the word relation and not equation in case of $\infty$ . The main thrust of my argument is how can you replace the expression under the root with x when you do not know if the expression is a real number at all??!!

My intent is to make you alive to these requirements and not go into it blindly.

I was taught this when I was coached for IIT. So, whether it was in the syllabus or not, I guess my teacher taught this for the sake of completeness. By the way, this is a problem in the opening chapter in Maron on sequences and it is very easy to follow the argument

I asked you guys to first try out

$x = \sqrt{2+\sqrt {2 + \sqrt {2+...\infty}}}$

Consider the series

$x_1 = \sqrt 2; x_2 = \sqrt {2+\sqrt 2}} ; x_3 = \sqrt {2+\sqrt {2+\sqrt 2}}} ...$

Obviously, this is an increasing sequence

Now, we have

$\sqrt 2 < 2 \Rightarrow \sqrt{2+\sqrt 2} \le \sqrt {2+2} = 2$

Now, $x_{n} = \sqrt{2+x_{n-1}}$ and hence $x_{n-1} < 2 \Rightarrow x_n = \sqrt{2+x_{n-1}} < \sqrt{2+2} = 2<br/>$

Thus the terms of the increasing sequence are bounded above and this means it is convergent to a real number.

It is at this stage that we can adopt the method used by rudra panda and others.

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