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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.

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

3 Answers 3

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$$ \begin{align} a_{n+1}-\frac34 &=\frac{a_n+3}5-\frac34\\ &=\frac{a_n-\frac34}5 \end{align} $$ Thus, if $a_n\ge\frac34$, then $a_{n+1}\ge\frac34$. Therefore, since $a_1=1$, $a_n\ge\frac34$.

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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.

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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.

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