$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

**Example**

Suppose that the sequence $a_n$ is defined by $a_1=0$ and $a_{n+1}=\sqrt{a_n+20}$ for all $n\in \N$. Show that $a_n$ is bounded and strictly increasing. Find its limit.

We need to find a bound. As $a_n$ is strictly increasing it will be bounded by its limit so perhaps we should find that first rather than find it last as asked in the question.

We have $a_{n+1}^2=a_n+20$ and so if $a_n\to a$ we have $a^2=a+20$ by the Algebra of Limits. (Don’t forget that $a_{n+1}$ and $a_n$ tend to the same limit.) As $a^2-a-20=(a+4)(a-5)$ we have $a=-4$ or $a=5$. Since $a_{n+1}$ is defined by a square root it is non-negative and hence $a=5$. Note that this shows that limit is $5$ *if* it exists. The if is important, we have not shown that the limit exists.

Since, for a bounded and increasing sequence the limit is equal to the smallest upper bound we know that the limit is a bound. Thus, if $5$ really is a limit, then it will be a bound.

The question tells us the sequence is increasing so let us prove that $5$ is an upper bound. That is, we claim that $a_n\leq 5$ for all $n\in \N$.

Since $a_n$ is defined inductively it makes sense to use induction. The initial case $n=1$ clearly holds. Thus, assume $a_k\leq 5$ for some $k\in \N$. Then,

\begin{align*}

a_{k+1}&=\sqrt{a_k+20}\\

&\leq \sqrt{5 +20} , {\text{ by the inductive hypothesis}},\\

&=5.

\end{align*}

Therefore, $a_n\leq 5$ for all $n\in \N$.

Next we look at the strictly increasing property. We have a connection between $a_{n+1}$ and $a_n$ in the definition of the sequence so it makes sense to start there and remove terms that have nothing to do with $a_n$ and $a_{n+1}$. We have

\begin{align*}

a_{n+1}&=\sqrt{a_n+20}, {\text{ we dislike the $20$, so replace with $a_n$ and gather terms,}} \\

&=\sqrt{a_n+4\cdot 5}\\

&\geq \sqrt{a_n+4a_n}\\

&=\sqrt{5a_n}, {\text{ we don’t like the $5$ so replace it}}, \\

&\geq \sqrt{a_na_n} ,{\text{ looking good!}},\\

&=a_n.

\end{align*}

Thus, $a_n$ is increasing. There is a problem though as we had to show that the sequence is *strictly* increasing! If the $\geq $ signs in the preceding were $>$ signs, then we would be ok. Thus we need to show that $a_n\lt 5$ rather than $a_n\leq 5$. The argument involving $a_n\leq 5$ will still hold true if we replace $\leq $ everywhere with $\lt $. Hence, we can show that $a_n$ is strictly increasing.

Let’s now give the polished version which will include this change to strict inequalities.

Note that we leave the calculation of the limit until the end, i.e., until after we prove that the sequence has a limit.

**Polished answer:**

First we show $a_n$ is strictly bounded by $5$, i.e., $a_n\lt 5$ for all $n\in \N$. By induction we have

\[

a_{n+1}=\sqrt{a_n+20}

\lt \sqrt{5 +20}

=5.

\]

Therefore, $a_n\lt 5$ for all $n\in \N$.

The sequence in strictly increasing:

\[

a_{n+1}

=\sqrt{a_n+4\cdot 5}

>\sqrt{a_n+4a_n}

=\sqrt{5a_n}

>\sqrt{a_na_n}

=a_n.

\]

As the sequence is bounded and increasing it converges to $a$ say. Thus,

\begin{align*}

a_{n+1}^2&=a_n+20\\

\lim_{n\to \infty} a_{n+1}^2&= \lim_{n\to \infty} (a_n+20)\\

a^2&=a+20, {\text{ by the Algebra of Limits}},\\

a^2-a-20&=0\\

(a+4)(a-5)&=0.

\end{align*}

As $a_1=0$ and $a_n$ is increasing we have $\displaystyle \lim_{n\to \infty} a_n=5$.