Series Convergence What does this series converge to? 
$$ \sqrt{3\sqrt {5\sqrt {3\sqrt {5\sqrt \cdots}}}} $$
and also this?
$$ \sqrt{6+\sqrt {6+\sqrt {6+\sqrt {6+\sqrt \cdots}}}} $$
And, generally speaking, how should one approach these kind of questions?
 A: Here's for
$$
\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\cdots}}}}.
$$
First define a sequence recursively by $a_1 := \sqrt{6}$ and $a_n := \sqrt{6+a_{n-1}}$ ($n \geq 2)$. Then we wish to compute $\lim_{n \to \infty} a_n$. In order to find this limit, we need, of course, to know that it exists. But instead of trying to show right away that it does, we will assume that it does and try to obtain candidates for its value. This will help us to show convergence (as it is often the case... in fact, most of the time, when we want to show that a sequence converges to some limit $L$, we need to first guess the value of $L$, which is not always simple... Cauchy sequences are useful justly because we don't have to guess a limit in order to prove convergence).
So, assuming the limit exists, put $\lim_{n \to \infty} a_n := L$. Then by definition of $a_n$ we have
$$
\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{6+a_n},
$$
which gives
$$
L = \sqrt{6+L}.
$$
In this case W.W. showed that $L \in \{-2, 3\}$ and correctly observed that $L = -2$ is impossible, since $a_n \geq 0$ $\forall n$. Thus, the only candidate for $L$ is $3$.
Now it is easily seen that $a_{n+1} \geq a_n$, so the sequence is increasing. In order to show convergence we know it suffices to exhibit an upper bound for $a_n$. But we might as well try to show that $a_n \leq 3$, in view of our candidate for $L$. This we can do by induction :


*

*Base step : Clearly $a_1 = \sqrt{6} \leq \sqrt{9} = 3$ is true.

*Induction step : Assume $a_{n-1} \leq 3$ for some $n \geq 2$. Then $a_n = \sqrt{6+a_{n-1}} \leq \sqrt{6+3} = 3$.


By the principle of mathematical induction, we have shown that $a_n \leq 3$ $\forall n$, which is what we were seeking.
We may conclude that $a_n$ converges and that its limit is 3.
A: Well, define
$$x=\sqrt{3\sqrt{5\sqrt3\ldots}}=\sqrt{3\sqrt{5x}}\stackrel{\text{square}}\implies x^2=3\sqrt{5x}\stackrel{\text{square again}}\implies x^4=45x\implies$$
$$x^3=45\iff x=\sqrt[3]{45}$$
But...the above follows at once from arithmetic of limits, on the sequence
$$\left\{\sqrt3\,,\,\sqrt{3\sqrt5}\,,\,\sqrt{3\sqrt{5\sqrt3}}\,\ldots\right\}$$
and we must first prove the above sequence converges finitely, so
Hints:
== Prove your sequence is monotone ascending
== Prove your sequence is bounded above
A: For $$\sqrt{6 + \sqrt{6 + \sqrt{6 +\dots}}}:$$
Let \begin{align*}
x &= \text{the given equation}\\
&= \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}
\end{align*}
Since the series is infinite, we can write
$x = \sqrt{6 + x}$,
or  $$x^2 - x - 6 = 0.$$
Therefore, $x = 3$ or $x = -2$
Since the answer cannot be negative, reject $x = -2$. Therefore, $x = 3$.
Do the same thing for the first series.
