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 $n \geq 1$, $$a_{n+1}>a_{n}.$$
The sequence appears to be increasing, and possibly bounded at 5. How may I show convergence, and find its limit.
Additionally are there any beginners guide on using appropriate notations?
Thank you for your assistance.
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 $a_{n+1}\le 5$.
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.
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 $a_{n+1}>a_n$.
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.
