# 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. • Do you mean "decreasing" instead of "increasing"? If so, your solution is good. Jan 11, 2020 at 19:31 •$(3/4 + 3)/5=3/4$Jan 11, 2020 at 19:31 • Instead of "$=15\gt\frac34$", you should have "$=\frac{15}{12}=\frac34\$"
– robjohn
Jan 11, 2020 at 19:36

\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 $$a_1=1$$, $$a_n\ge\frac34$$.
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.
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 $$f(x) (decreasing sequence).
Now the defining function is here $$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.