Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or  $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$ 
 I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But I don't see a pattern in the left side.
 A: If you do repeated squaring and subtracting what's outside the square root, the left-hand side (the one starting at $3$) will follow the pattern
$$
3\\
3^2-2=7\\
7^2-3=46\\\vdots
$$
It may be difficult to find some explicit form for an arbitrary term here. But we don't need that. We only need to show that term number 100 (i.e. after squaring and subtracting $100$) is positive. And that allows us to do massive simplifications. It is, for instance, easy enough to show that the sequence more than doubles each step, which trivially implies that it is always strictly positive.
Let $a_n$ denote the $n$th entry in the sequence above. We have $a_1=3$ and
$$
a_n=a_{n-1}^2-n
$$
for all $n>1$. We see that in the first step, $a_2=7$ is more than double of $a_1=3$.
Take $n\geq 2$, and assume $a_n>2^{n-1}\cdot3$. Then
$$
a_{n+1}=a_n^2-n-1=(a_n -1)a_n+(a_n-n-1)\\
> 2a_n+0
$$
which by induction proves that the sequence more than doubles each step. Here I have used $a_n>3$ and $a_n\geq n$, both of which may be deduced from $a_n>2^{n-1}\cdot3$.
A: Intuitively, all those big numbers at the right are under so many square root signs they don't matter much.  I would think it "obvious" that $3$ is larger.  I made a spreadsheet (love copy down) and found the value to be about $2.0903$.  For comparison, I changed all the numbers after $9$ to $0$ and the value only changed in the sixth decimal place.  
Proving it without computer calculation is another matter.  One approach is just to unpack the square roots one by one.  We want to compare $9$ with ${2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}}$, which means comparing $7$ with $ \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}$, which means comparing $49$ with $3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46$ with $ \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46^2$ with $4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}$ and now we can afford to take all the square root signs except the first off the right and still have the left larger- $46^2 \gt 4+\sqrt{5+6+7+8+\ldots 100}=4+\sqrt{5044}$.
A: Let $x$ be the given iterated radical.  Render
$x^2=2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
Now multiply $x$ by $\sqrt{3/2}$:
$(\sqrt{3/2})x=\sqrt{3+(3/2)\sqrt{4+\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+(9/4)\sqrt{5+\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+(81/16)\sqrt{6+...}}}}$
$=\sqrt{3+\sqrt{9+\sqrt{(405/16)+\sqrt{(19683/128)+...}}}}$
The second and subsequent radicands in $(\sqrt{3/2})x$ are larger than those in $x^2-2$ forcing
$x^2-2<(\sqrt{3/2})x$
$x^2-(\sqrt{3/2})x-2<0$
$x<\sqrt{3/8}+\sqrt{19/8}$ from the quadratic formula
The sum of two square roots is less than twice the square root of the average of the radicands:
$x<2\sqrt{11/8}, x^2<11/2<9, \color{blue}{x<3}$
A: Actually, there is a rather simple method to do this:
We can rewrite 
$$a=\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$$
into 
$$b=\sqrt{2 + \sqrt{3 + \sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}}}$$
And we can clearly see that
$$b>a$$
There is a classic method to evaluate continued root. Let 
$$x=\sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}$$
Then 
$$x=\sqrt{100+x}$$
$$x^2=100+x$$
$$x=\frac{1+\sqrt{401}}{2}$$
Obviously,
$$20<\sqrt{401}<21$$
We see that
$$x<\frac{1+21}{2}=11$$
Then we have 
$$b<\sqrt{2 + \sqrt{3+11}}$$
$$b<\sqrt{2 + \sqrt{14}}<\sqrt{6}<3$$
So
$$a<b<3$$
