Proving that the sequence $a_1 = 4$, $a_{n+1} = \sqrt{a_{n} +20}$ converges and finding its limit 
Given a sequence $\{a_n\}$, with $n\geq1$, where
  $$a_{1}=4,$$
  and 
  $$a_{n+1}=\sqrt{a_{n} +20}.$$
  Prove via induction that, for all $n \geq 1$, 
   $$a_{n+1}>a_{n}.$$



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*Apparent Convergence


The sequence appears to be increasing, and possibly bounded at 5. How may I show convergence, and find its limit.


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*Regarding Notations


Additionally are there any beginners guide on using appropriate notations?
Thank you for your assistance.
 A: First let's guess what the limit is (if exists): say $a_n\to\alpha$, then, by continuity of the operations $+$ and $\sqrt{\ \ }$, we have
$$\alpha = \sqrt{\alpha+20},\ \text{  i.e., }\  \alpha^2=\alpha+20$$
Its roots are $-4$ and $5$. Since $a_n>0$ always, only $\alpha=5$ can be valid.
Now use induction, with additional hypothesis that $a_n\le 5$ to prove $a_{n+1}>a_n$ and still $a_{n+1}\le 5$.
A: The induction proofs write themselves. We show first that $a_{n+1}\gt a_n$ for all $n$. The result is true at $n=1$. Suppose that we know that $a_{k+1}\gt a_k$ for some particular $k$.  We need to show that $a_{k+2}\gt a_{k+1}$. This is straightforward: 
$$a_{k+2}=\sqrt{a_{k+1}+20}\gt \sqrt{a_k+20}=a_{k+1}.$$
We prove, again by induction, that $a_n\lt 5$ for all $n$. This is true at $n=1$. Suppose that we know that $a_k\lt 5$. We show that $a_{k+1}\lt 5$. This is easy:
$$a_{k+1}=\sqrt{a_k+20}\lt \sqrt{25}=5.$$
The sequence $(a_n)$ is increasing and bounded above by $5$, so it has a limit. That the limit is $5$ is proved as in the answer by Berci Pecsi. 
A: EDIT: This is still false. For example, the sequence $4, 10, 10, 10, 10, \ldots$ satisfies $a_{n+1}>\sqrt{a_n+20}$ for all $n$, and $a_1=4$, but not $a_{n+1}>a_n$.
A: Consider another sequence $b_{n}$
  with the same definition 
$$b_{n+1}=\sqrt{b_{n}+20}$$
Since the function $f(x)=\sqrt{x+20}  $ is an increasing function, 
If $b_{1}>a_{1}$
 , then obvioulsy $b_{2}>a_{2}$
  and so on an $b_{n}>a_{n}$
  for all $n$
 ,
Let $b_{1}=5$
 , then $b_{n}=5$
  for all n.
  Since $b_{1}>a_{1}$
 , we have $b_{n}>a_{n}\implies5>a_{n}$
 .
Since $a_{n}$
 is monotonically increasing and has an upper bound, it must converge.
