show that the following recursive series is bounded and monotonic I need to show that the following recursive series $$a_1=1, \quad a_{n+1}=\frac{a_n+3}{5}$$ is bounded and monotonic.
At first, I want to show that the series is bounded from below with $\frac{3}{4}$. By induction:
For $n=1$ we have $a_1=1>\frac{3}{4}$. We assume it is true for $n$, i.e. $a_n>\frac{3}{4}$ and we check for $a_{n+1}$:
$$a_{n+1}=\frac{a_n+3}{5}>\frac{\frac{3}{4}+3}{5}=15>\frac{3}{4}.$$
We conclude that $a_n>\frac{3}{4}$ for all n.
The series is increasing because
$a_{n+1}-a_{n}=\frac{-4a_n+3}{5}=\frac{-4(a_n-\frac{3}{4})}{5}<0$$
so we have also $a_n \leq a_1=1$.
Is my solution ok? Thank you for your help.
 A: $$
\begin{align}
a_{n+1}-\frac34
&=\frac{a_n+3}5-\frac34\\
&=\frac{a_n-\frac34}5
\end{align}
$$
Thus, if $a_n\ge\frac34$, then $a_{n+1}\ge\frac34$. Therefore, since $a_1=1$, $a_n\ge\frac34$.
A: You can solve the $a_n$. Actually, for $a_{n+1}=\frac{a_n+3}5$
$$\therefore a_{n+1}+k=\frac{a_n+3+5k}5 $$
Make $k=3+5k$, so $k=-\frac 34$, now
$$a_{n+1}-\frac 34=\frac{a_n+3-\frac{15}4}5=\frac{a_n-\frac 34}5 $$
Let $b_n=a_n-\frac 34$, now $b_1=\frac 14$, and
$$b_{n+1}=\frac{b_n}5$$
So we know $a_n-\frac 34=b_n=\frac 1{4\cdot 5^{n-1}}$, so $a_n=\frac 34+\frac 1{4\cdot 5^{n-1}}$, it is bounded and monotonic.
A: With very little computations:
To show a recursive sequence is monotone, you just have to show all its terms belong to an interval $I$ on which $f(x)>x$ (increasing sequence) or $f(x)<x$ (decreasing sequence).
Now the defining function is here $f(x)=\smash[b]{\dfrac{x+3}5}$, and its graph is above the line $y=x$ if $x<\frac34$, below if  $x>\frac 34$.
On the other, it is easy to see that $f$ maps the interval $\;I=\bigl[\frac34,+\infty\bigr)$ onto itself. As $a_i\in I$, each $a_n\in I$ and the sequence decreases. As it is bounded from below by, say $\frac 34$, it converges.
