Solutions added

Finally added some solutions to the problems by updating the problem post itself. 

If there are any missing which you would like to see, please comment on the problem post.

A surprising expression with nested square roots

 Simplify 

$$ \frac{\sqrt{10 + \sqrt{1}} + \sqrt{10 + \sqrt{2}} + \dots + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \dots + \sqrt{10 - \sqrt{99}}}$$

Scroll down for simplified form and solution.

The expression is equal to $\sqrt{2} + 1$.  Surprising!

Scroll down for solution.

Let $a + b $ = 100. 

Now

$$\sqrt{10 - \sqrt{a}} \sqrt{10 + \sqrt{a}} = \sqrt{100 - a}  = \sqrt{b}$$

Let $u = \sqrt{10 + \sqrt{a}}$ and $v = \sqrt{10 - \sqrt{a}}$

So $20 = u^2 + v^2$

and $\sqrt{b} = uv$

Thus

$$\sqrt{20 + 2\sqrt{b}} = \sqrt{u^2 + v^2 + 2uv} = u+v$$

Similarly

$$\sqrt{20 - 2\sqrt{b}} = \sqrt{u^2 + v^2 - 2uv} = u-v$$

Thus if 

$$P = \sum_{a=1}^{99} \sqrt{10 + \sqrt{a}}$$

and

$$Q = \sum_{a=1}^{99} \sqrt{10 - \sqrt{a}}$$

We get 

$$ \sqrt{2} P = P + Q$$ and

$$\sqrt{2} Q = P - Q$$

This gives us $$\frac{P}{Q} = \sqrt{2} + 1$$

Putnam 2012 B4 generalization

 Let $c \gt 0$ be a real number and $a_n$ be a sequence such that $a_0 = 1$ and $$a_{n+1} = a_n  + e^{-c a_n}$$

Show that

$$\lim_{n \to \infty} (ca_n - \log n) = \log c$$

($\log$ is $\log$ to base $e$)

The Putnam problem was with $c = 1$ and only asked for proof of existence of the limit.

Scroll down for a solution.

Consider $b_n = e^{a_n}$ then we get that $b_{0} = e$ and

$$ b_{n+1} = b_n e^{1/(b_n)^c}$$

We can easily show that $a_n$ is unbounded (proof by contradiction) and so is $b_n$ and thus $\frac{1}{b_{n}^c} \to 0$.

The recurrence for $a_n$ gives us

$$b_{n+1}^c = b_{n}^c e^{c/(b_n)^c}$$

Expanding the $e^{\dots}$ part we get

$$b_{n+1}^c = b_{n}^c ( 1 + \frac{c}{b_{n}^c} + O\left(\frac{1}{b_{n}^{2c}}\right)) = b_{n}^c + c + O\left(\frac{1}{b_{n}^c}\right)$$

This telescopes to give us

$$b_{n}^c - b_{0}^c = nc + \sum_{k=0}^{n} O\left(\frac{1}{b_{k}^c}\right)$$

And so

$$\frac{b_{n}^c - b_{0}^c}{n} = c + \frac{\sum_{k=0}^{n} O\left(\frac{1}{b_{k}^c}\right)}{n}$$

Thus

$$ \frac{b_{n}^c}{n} \to c$$

Taking logarithms gives the result.