How to show that the following sequence is monotinacly increasing?

Question

given $$a_{1}=1, \quad a_{n+1}=\frac{2\left(2 a_{n}+1\right)}{a_{n}+3}$$ I am trying to show $a_{n}\leq a_{n+1}$ I am trying to show in induction, so my assumption is that $a_{n}\leq a_{n+1}$ and I want to show $a_{n+1}\leq a_{n+2}$ the only thing I could get is that $$a_{n+2}=\frac{2\left(2 a_{n+1}+1\right)}{a_{n+1}+3}>\frac{2\left(2 a_{n}+1\right)}{a_{n+1}+3}=a_{n+1}\cdot \frac{(a_n +3)}{a_{n+1}+3}$$

Answer 1

Let $ f(x) = \frac{ 2 (2x+1) } { x+3}$.

Hint: Show that $ f(x)$ is an increasing function on (at least) $x \geq 1$.

Using Nightflight's hint of $ f(x) = 4 - \frac{10}{x+3}$, we can avoid using calculus.

Corollary: The induction hypothesis follows. Since $a_n \leq a_{n+1}$, hence $f(a_n) \leq f(a_{n+1} )$.

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Simplification of Eric's solution.

Hint: Show that $ x \leq f(x) \leq 2$ on (at least) $[1,2]$.

Corollary: Using induction, prove the stronger statement that $1 \leq a_n \leq a_{n+1} \leq 2$.

Answer 2

Notice that $a_n$ had better never be $-3$ or the recursion fails to be defined. We wish to show \begin{align*} a_n &\leq a_{n+1} \\ &= \frac{2(2a_n+1)}{a_n+3} \\ &= \frac{4a_n+2}{a_n+3} \end{align*} so \begin{align*} \begin{cases} 4a_n + 2 \geq a_n(a_n+3) ,& a_n+3 > 0 \\ 4a_n + 2 \leq a_n(a_n+3) ,& a_n+3 < 0 \end{cases} \\ \begin{cases} 4a_n + 2 \geq a_n^2+3a_n ,& a_n > -3 \\ 4a_n + 2 \leq a_n^2+3a_n ,& a_n < -3 \end{cases} \\ \begin{cases} 0 \geq a_n^2-a_n-2 ,& a_n > -3 \\ 0 \leq a_n^2-a_n-2 ,& a_n < -3 \end{cases} \\ \begin{cases} 0 \geq (a_n - 2)(a_n + 1) ,& a_n > -3 \\ 0 \leq (a_n - 2)(a_n + 1) ,& a_n < -3 \end{cases} \\ \end{align*} Drawing sample points from $(-\infty,-3)$, $(-3,-1)$, $(-1,2)$, and $(2,\infty)$, we find that this is satisfied for $a_n < 3$ and $a_n \in (-1,2)$.

We have $a_1 = 1$. To be monotonically increasing, we must have $a_n \in [1,2)$ for all $n \geq 1$. (In particular, although we get at least one step of increasing if $a_n < -3$, if any $a_n$ after $a_1$ is less than $a_1 = 1$, the sequence can't be increasing.) We have $\frac{\mathrm{d}}{\mathrm{d}x} \frac{2(2x+1)}{x+3} = \frac{10}{(x+3)^2} > 0$ (for $x \neq -3$), so is increasing on $[1,2)$. This means $\frac{2(2x+1)}{x+3}$ is lower bounded at $x = 1$ and is upper bounded at $x = 2$. We compute $\frac{2(2(1)+1)}{(1)+3} = \frac{3}{2}$ and $\frac{2(2(2)+1)}{(2)+3} = 2$, so $a_n \in [3/2,2)$ for all $n$. Therefore, the sequence is monotonically increasing.

Answer 3

If you want even more, look at my answer to this question.

Applied to your simple case (simple because of the specific cefficients), this will give $$a_n=2-\frac{15\ 2^n}{5\ 2^n+4\ 5^n}$$