Monotone and Bounded Sequences: Proof Say $s_1=1$ and $s_{n+1}=\frac{1}{5}(s_n+7)$ for $n\geq 1$. Prove that the sequence is monotone and bounded, then find the limit. 
 A: We have
$$
s_{n+1}-s_n=\frac15(s_n-s_{n-1}) \quad \forall n \ge 2,
$$
and so we have
$$\tag{*}
s_{n+1}-s_n=\frac{s_2-s_1}{5^{n-1}}=\frac{3}{5^n} \quad \forall n \ge 1. 
$$
Thus $s_n<s_{n+1}$ for every $n \ge 1$, i.e. $(s_n)$ is increasing.
Thanks to (*) we have for every $n \ge 1$:
\begin{eqnarray}
s_n&=&1+\sum_{i=1}^{n-1}(s_{i+1}-s_i)=1+\sum_{i=1}^{n-1}\frac{3}{5^i}=-2+3\sum_{i=0}^{n-1}\frac{1}{5^i}=-2+3\cdot\frac{1-\frac{1}{5^n}}{1-\frac15}\\
&=&-2+\frac{15}{4}\left(1-\frac{1}{5^n}\right)=\frac74-\frac34\cdot\frac{1}{5^{n-1}}<\frac74.
\end{eqnarray}
We deduce that $(s_n)$ is a convergent sequence and 
$$
\lim_ns_n=\lim_n\left(\frac74-\frac34\cdot\frac{1}{5^{n-1}}\right)=\frac74.
$$
A: $$\begin{align*}\bullet&\;\;s_1=1\le\frac{1+7}5=:s_2\;,\;\;\text{induction now:}\\
s_n:&=\frac15(s_{n-1}+7)\stackrel{\text{ind. hyp.}}\le\frac15(s_n+7)=:s_{n+1}\\{}\\
\bullet\bullet&\;\;\text{Bounded by induction again:}\;\;s_1=1\le 2\;,\;\;s_{n+1}=\frac15(s_n+7)\le\frac15(2+7)\le 2\end{align*}$$
Thus the limit exists, call it $\;\alpha\;$ , and using arithmetic of limits we get
$$\alpha\leftarrow s_n=\frac15(s_n+7)\to\frac15(\alpha+7)\implies5\alpha=\alpha+7\ldots$$
and the limit indeed is what you think it is
A: We begin with the limit. Suppose that 
$$\lim_{n \rightarrow \infty} (s_n)= L.$$
Now
\begin{align*}
& s_{n+1}=\frac{1}{5}(s_n+7) \\
\Rightarrow & \lim_{n \rightarrow \infty} s_{n+1} = \lim_{n \rightarrow \infty} \frac{1}{5}(s_n+7) \\
\Rightarrow & L = \frac{1}{5}(L+7) \\
\Rightarrow & \frac{4}{5}L =\frac{7}{5} \\
\Rightarrow & L =\frac{7}{4}.
\end{align*}
We conclude that
$$\lim_{n \rightarrow \infty}(s_n) = \frac{7}{4}.$$
$\textbf{Claim:}$ The sequence $(s_n)$ is bounded so that $1 \leq s_n < \frac{7}{4}$.
$\textbf{Proof:}$
Let $n=1$. Then $s_1=1$, so $1 \leq s_1 < \frac{7}{4}$. Assume that the sequence is bounded for $n=k$. Then $1 \leq s_k < \frac{7}{4}$. Now
\begin{align*}
& 1 \leq s_k < \frac{7}{4} \\
\Rightarrow & 8 \leq s_k +7 < \frac{35}{4} \\
\Rightarrow & \frac{8}{5} \leq \frac{1}{5}(s_k+7) < \frac{7}{4} \\
\Rightarrow & \frac{8}{5} \leq s_{k+1} < \frac{7}{4}.
\end{align*}
Since $\frac{8}{5} \leq s_{k+1} < \frac{7}{4}$, certainly $1 \leq s_{k+1} < \frac{7}{4}$, and so we conclude that $(s_n)$ is bounded for all $n \in \mathbb{N}$.
$\textbf{Claim:}$ The sequence $(s_n)$ is monotone increasing so that $s_{n+1}-s_n > 0 \, \forall \, n \in \mathbb{N}$.
$\textbf{Proof:}$
We take the difference.
\begin{align*}
s_{n+1}-s_n &= \frac{1}{5}(s_n+7)-s_n \\
            &= \frac{7}{5}-\frac{4s_n}{5} \\
            &= \frac{7-4s_n}{5}.
\end{align*}
Now 
$$1 \leq s_n < \frac{7}{4} \Rightarrow  7-4s_n > 0 \Rightarrow \frac{7-4s_n}{5}>0 \Rightarrow  s_{n+1}-s_n > 0 \, \forall \, n \in \mathbb{N}.$$
We conclude that the sequence $(s_n)$ is monotone increasing.
Now you are clearly using the property that every bounded monotone sequence converges to show that the sequence converges to the limit. You might want to put your limit proof last. I put it first just so we could use it in our proofs of monotonicity and boundedness.
