Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$ I know the sequence is converging. But I find it difficult proving it, by induction. So far I have drawn a diagram and calculate the five first numbers. From the diagram I can se that the sequence can be split into two sequences, one that is increasing and one tha is decreasing. I need to show that 

I need to show that
  $a_{{1}}=1$,
  $a_{{3}}=3/2$ and
  $a_{{5}}={\frac {30}{17}}$ is an increasing sequence and that
  $a_{{2}}=3$, 
  $a_{{4}}={\frac {12}{5}}$ is decreasing. 

Any help would be appriciated. 
$a_{{1}}=1, a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1},1\leq n$


*

*$a_{{1}}=1$

*$a_{{2}}=3$

*$a_{{3}}=3/2$

*$a_{{4}}={\frac {12}{5}}$

*$a_{{5}}={\frac {30}{17}}$

 A: (At the end,
I have added my form
of the explicit solution.)
$a_{1}=1, a_{n+1}=\dfrac{6}{a_{{n}}+1},1\leq n
$
If it has a limit $L$,
then
$L = \dfrac{6}{L+1}$
or
$L^2+L-6 = 0
$
or
$(L+3)(L-2) = 0
$.
Since
$a_n > 0$
for all $n$,
the only possibility
is $L=2$.
To get a sequence
whose terms should
go to $0$,
let
$a_n = b_n+2
$.
Then
$b_{n+1}+2=\dfrac{6}{b_{{n}}+3}
$
or
$b_{n+1}
=\dfrac{6-2(b_{n}+3)}{b_{n}+3}
=\dfrac{-2b_{n}}{b_{n}+3}
$.
Since
$b_1 = a_1-2
= -1
$,
$b_2 = \dfrac{2}{2}
= 1
$,
$b_3 = \dfrac{-2}{4}
= \dfrac12
$,
$b_4 = \dfrac{-1}{7/2}
= -\dfrac{2}{7}
$.
Since
$\left(-\dfrac{2 x}{x+3}\right)' = -\dfrac{6}{(x+3)^2}
$,
if
$-\dfrac12 \le b_n
\le \dfrac12$,
then
$b_{n+1}
\le \dfrac{1}{5/2}
=\dfrac25
\lt \dfrac12
$
and
$b_{n+1}
\ge \dfrac{-1}{7/2}
=-\dfrac27
\gt -\dfrac12
$.
Therefore
$|b_n|
\le \dfrac12
$
for $n \ge 3$.
Therefore,
for
$n \ge 3$,
$\begin{array}\\
\left|\dfrac{b_{n+1}}{b_n}\right|
&=\left|\dfrac{2}{b_n+3}\right|\\
&\le \dfrac{2}{5/2}\\
&= \dfrac45\\
&< 1\\
\end{array}
$
so that
$b_n
\to 0
$
as $n \to \infty$.
Therefore
$a_n
\to 2
$
as $n \to \infty$.
(added later)
I can show that
$b_m
= \frac{15}{8(-\frac32)^m-3}
$
so that
$b_{2m}
=\frac{15}{8(\frac94)^{m}-3}
$
is decreasing
and
$b_{2m+1}
=\frac{-5}{4(\frac94)^{m}+1}
$
is increasing,
both approaching $0$
as their limit.
A: You can actually outright solve this sequence and take the limit. You can prove this by induction.
$$a_n = \frac{15 (-2)^n}{8 \times 3^n - 3(-2)^n} + 2$$ 
Taking the limit as $n \to \infty$ gives you 2.
