Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$

Question

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.

$a_{{1}}=1, a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1},1\leq n$

• $a_{{1}}=1$
• $a_{{2}}=3$
• $a_{{3}}=3/2$
• $a_{{4}}={\frac {12}{5}}$
• $a_{{5}}={\frac {30}{17}}$

Answer 1

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.

Answer 2

(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.