Prove that sequence is monotone with induction

Question

$$a_{n+1} = \frac{2a_n}{3+a_n}, a_1=1$$

I was trying to solve it by using induction.

I assumed that $a_{k+1}<a_k$ and tried to to prove that $a_{k+2}<a_{k+1}$.

First I tried to use what I assumed and "build" it to what I need to prove. It didn't work so I started again with $a_{k+2}<a_{k+1}$ and tried to build it to $a_{k+1}<a_k$. I couldn't make it either way.

Answer 1

Because $a_n>0$ and $$a_{n+1}-a_n=\frac{2a_n}{3+a_n}-a_n=\frac{-a_n^2-a_n}{3+a_n}<0$$

Answer 2

It is obviously that $a_n>0$ for all $n$. We prove that the sequence is decreasing. Suppose there is $n$ such that $a_n\leq a_{n+1}$, so we have $$ {2a_n\over 3+a_n} \geq a_n $$ which is equivalent to $ -1\geq a_n$. A contradiction.

Answer 3

If $a_n>0$ then $a_{n+1}>0$ and since $a_0=1$ then $a_n>0$ for all $n$.

$0<\dfrac {a_{n+1}}{a_n}=\dfrac 2{3+a_n}<\dfrac 23<1$

So $a_{n+1}<a_n$

Answer 4

Beware: overkill. I am going to compute an explicit form for $a_n$. By setting $a_n=\frac{p_n}{q_n}$ with $p_1=q_1=1$, $v_n=(p_n\,q_n)^T$ and $M=\begin{pmatrix}2 & 0 \\ 1 & 3\end{pmatrix}$ we have $$ v_{n+1} = M^{n} v_1.$$ The eigenvalues of $M$ are $2,3$ and by the Cayley-Hamilton theorem the sequences $\{v_n\}_{n\geq 1},\{p_n\}_{n\geq 1},\{q_n\}_{n\geq 1}$ and the matrix $M$ have the same characteristic polynomial.
In particular

$$a_n = \frac{A\cdot 3^n+B\cdot 2^n}{C\cdot 3^n+D\cdot 2^n}$$ where the constants $A,B,C,D$ can be found by interpolation, given $p_1=q_1=1$, $p_2=2$, $q_2=4$. In explicit terms $$ a_n = \frac{2^{n-1}}{2\cdot 3^{n-1}-2^{n-1}}=\frac{1}{2\left(\frac{3}{2}\right)^{n-1}-1} $$ is pretty obviously decreasing and convergent to zero, since $2\left(\frac{3}{2}\right)^{n-1}-1$ is increasing and divergent to $+\infty$.