Convergence and Limit of a Recursive Sequence from the multiples of $a_n$ I'm having trouble with this recursive sequence problem. I'm supposed to find the limit, assuming that is it convergent, but I can't seem to get the answer.
$a_1 = 1, a_{n+1}= \frac {2a_n}{7+a_n}  $ 
 A: If a sequence is convergent then we know that $\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a_n = L$. take the limit of both sides and you will get:
$$\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{2a_n}{7+a_n} = \frac{2 \cdot \lim_{n \to \infty} a_{n}}{7 + \lim_{n \to \infty} a_n} \implies L = \frac{2L}{7+L}$$
This should be an easy quadratic equation. Upon solving it you will get 2 possibilities and you can easily discard one of them.

To see why it's convergent note that the sequence is bounded from below by $0$ and it's decreasing, as $a_{n+1} = \frac{2a_n}{7+a_n} < a_n \iff 2a_n < 7a_n + a_n^2 \iff 0 < 5a_n + a_n^2$. You can sort all the technicalities using induction. As it's decreasing and bounded from below the sequence is convergent.
A: We may just find the explicit form of $a_n$. By setting $a_n=\frac{p_n}{q_n}$ we have:
$$\begin{pmatrix} p_{n+1} \\ q_{n+1}\end{pmatrix}=\begin{pmatrix} 2 & 0 \\ 1 & 7\end{pmatrix}\begin{pmatrix} p_{n} \\ q_{n}\end{pmatrix}$$
where the characteristic polynomial of the involved matrix is $(x-2)(x-7)$.
By the Cayley-Hamilton theorem it follows that:
$$ p_n = A\cdot 7^n + B\cdot 2^n,\qquad q_n = C\cdot 7^n + D\cdot 2^n $$
and since $p_1=q_1=1$ and $p_2=2,q_2=8$ we have:
$$ p_n = 2^{n-1},\qquad q_n = \frac{6}{35}\cdot 7^n-\frac{1}{10}\cdot 2^n $$
so:

$$ a_n = \color{red}{\frac{5}{6\cdot \left(\frac{7}{2}\right)^{n-1}-1}}$$

and $\lim_{n\to +\infty} a_n = \color{red}{0}$ trivially follows.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{a_{1} = 1}$:
$$
\mbox{Note that}\quad
a_{n + 1} = {2a_{n} \over 7 + a_{n}}
\quad\imp\quad
{1 \over a_{n + 1}} = {7 \over 2}\,{1 \over a_{n}} + \half
\quad\imp\quad
{1 \over a_{n + 1}} + {1 \over 5} = {7 \over 2}\,\pars{{1 \over a_{n}} +
{1 \over 5}}
$$

Then,
\begin{align}
{1 \over a_{n}} + {1 \over 5} & =
{7 \over 2}\,\pars{{1 \over a_{n - 1}} + {1 \over 5}} =
\pars{7 \over 2}^{2}\,\pars{{1 \over a_{n - 2}} + {1 \over 5}} = \cdots =
\pars{7 \over 2}^{n}\,\pars{{1 \over a_{1}} + {1 \over 5}} =
\pars{7 \over 2}^{n - 1}{6 \over 5}
\\[3mm] \imp\quad &
\color{#f00}{a_{n}} = \color{#f00}{{5 \over 6\pars{7/2}^{n - 1} - 1}}
\quad\imp\quad
\color{#f00}{\lim_{n \to \infty}a_{n}} = \color{#f00}{0}
\end{align}
A: Let define the following function: $$f:x\mapsto\frac{2x}{7+x}.$$
Some observations $f([0,+\infty[)\subseteq[0,+\infty[$ and $f$ is an increasing function on $[0,+\infty[$. Therefore, $(a_n)_n$ is an increasing sequence. Therefore, $(a_n)_n$ is convergent if and only if $(a_n)_n$ is bounded. Notice that if $(a_n)_n$ is convergent to $l$, $l$ can only be a fixed point of $f$ in $[0,+\infty[$ i.e. $f(l)=l$. Whence, if $(a_n)_n$ is convergent, $(a_n)_n$ is convergent to $0$.
