Nice exericse asked by Dan Sitaru in many math groups

Exercise:

Let ${{\left( {{a}_{n}} \right)}_{n\ge 1}}$ be recursive sequence defined in such way

${{a}_{1}}=a\in \mathbb{R}\backslash \mathbb{Q}$ and $2{{a}_{n}}{{a}_{n+1}}=4{{a}_{n}}+3{{a}_{n+1}}$ for all $n\ge 1$

1) Find the recursive closed formula for ${{a}_{n}}$

2) find$\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}$

Solution:

1) we have $2{{a}_{n}}{{a}_{n+1}}=4{{a}_{n}}+3{{a}_{n+1}}\Leftrightarrow \frac{4{{a}_{n}}+3{{a}_{n+1}}}{{{a}_{n}}{{a}_{n+1}}}=2\Leftrightarrow \frac{4}{{{a}_{n+1}}}+\frac{3}{{{a}_{n}}}=2$

notice that, $2=\frac{8+6}{7}=\frac{4\times 2}{7}+\frac{2\times 3}{7}$  , then

$4\left[ \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right]+3\left[ \frac{1}{{{a}_{n}}}-\frac{2}{7} \right]=0\Leftrightarrow \left( \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right)+\frac{3}{4}\left( \frac{1}{{{a}_{n}}}-\frac{2}{7} \right)=0$

Put ${{w}_{n}}=\frac{1}{{{a}_{n}}}-\frac{2}{7}\Leftrightarrow {{w}_{n+1}}=\frac{1}{{{a}_{n+1}}}-\frac{2}{7}\Leftrightarrow {{w}_{n+1}}=-\frac{3}{4}{{w}_{n}}$

Note that, for $n=1$ we have

                       $\begin{align}
  & {{w}_{2}}=\left( -\frac{3}{4} \right){{w}_{1}} \\
 & {{w}_{3}}=\left( \frac{-3}{4} \right){{w}_{2}} \\
 & \,\,\,\,\,\,\,\vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
 & {{w}_{n+1}}=\left( \frac{-3}{4} \right){{w}_{n}} \\
\end{align}$

Hence ${{w}_{n+1}}=\underbrace{\left( \frac{-3}{4} \right)\left( \frac{-3}{4} \right)....\left( \frac{-3}{4} \right)}_{n-time}{{w}_{1}}\Leftrightarrow {{w}_{n+1}}={{\left( \frac{-3}{4} \right)}^{n}}{{w}_{1}}\,\,\forall n\ge 1$

Therefore, $\left( \frac{1}{{{a}_{n+1}}}-\frac{2}{7} \right)={{\left( \frac{3}{4} \right)}^{n}}\left( \frac{1}{{{a}_{1}}}-\frac{2}{7} \right)\Leftrightarrow \frac{1}{{{a}_{k}}}={{\left( \frac{3}{4} \right)}^{k-1}}\left( \frac{1}{a}-\frac{2}{7} \right)+\frac{2}{7}\,\,\,\,\,\forall k\ge 2$

2) as $n\to \infty $ then ${{\left( \frac{-3}{4} \right)}^{n-1}}\to 0$  hence $\frac{1}{{{a}_{n}}}\to \frac{2}{7}$ thus $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=\frac{7}{2}=3.5$

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idea of solution credit to professor Ravi Prakash