Let $a_{n+1}=\frac{2a_{n}+1}{a_{n}+1}$ .For $a_{1}$=1 can we show that $a_{n}$<2

Question

I tried to prove first that $a_{n}$ is increasing but found a boundary for the value of $a_{n}$.So can i first show that $a_{n}$ <2 in order to overcome the problem of the boundary?

Answer 1

$$a_{n+1}=\frac{2a_n+2-1}{a_n+1}=2-\frac{1}{a_n+1}<2.$$ Also, $a_1<2$.

Answer 2

Sequences with such a structure always have a reasonably simple closed form. Set $a_n=\frac{p_n}{q_n}$ ($p_1=q_1=1$) and $v_n=(p_n,q_n)^T$. We can translate $a_{n+1}=\frac{2a_n+1}{a_n+1}$ first into $\frac{p_{n+1}}{q_{n+1}}=\frac{2p_n+q_n}{p_n+q_n}$ then into $$ v_{n+1} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} v_n^T = M v_n^T.$$ The eigenvalues of $M$ are $\frac{3\pm\sqrt{5}}{2}$ and the Cayley-Hamilton theorem ensures that $$ p_n = A\varphi^{2n}+B\overline{\varphi}^{2n},\qquad p_n = C\varphi^{2n}+D\overline{\varphi}^{2n} $$ for a set of constants $\{A,B,C,D\}$ which can be found by interpolation.
In our case we simply have $p_n=F_{2n}$ and $q_n=F_{2n-1}$ with $\{F_n\}_{n\geq 0}$ being the sequence of Fibonacci numbers, and $F_{2n}\leq 2 F_{2n-1}$ is straightforward to prove by induction.

Answer 3

$1<2$,
$2a_n+1<2a_n+2$ as long as $a_n$ are positive, so $(2a_n+1)/(a_n+1) < 2 $ and you are done.