Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence).
Let $a_n=\frac{b_{n+1}}{b_n}$ where $b_1=1, b_2=1$ and $b_{n+2}=b_n+b_{n+1}$ , Then $\lim\limits_{n \to \infty} a_n$ is
Real Analysis
Sequence of Reals
Limit of a Sequence
But try the problem first...
Answer: $\frac{1+\sqrt5}{2}$
IIT JAM 2018 (Problem 2)
Advanced Calculus by Patrick Fitzpatrick
First hint
Given that, $a_n=\frac{b_{n+1}}{b_n}$
$\Rightarrow \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \frac{b_{n+1}}{b_n}= \mathcal{L} $ (say)
Now we know that , $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} a_{n+1} $
$\Rightarrow \mathcal{L}=\lim\limits_{n \to \infty} a_{n+1}$
Can you find an equation on $\mathcal{L}$ from which the value of $\mathcal{L}$ can be obtained.
Second Hint
$\mathcal{L}= \lim\limits_{n \to \infty } a_{n+1}$
$= \lim\limits_{n \to \infty} \frac{b_{n+2}}{b_{n+2}}$
$=\lim\limits_{n\to \infty} \frac{b_{n+1}+b_n}{b_{n+1}}$ [By the given recurrence relation]
$=\lim\limits_{n\to \infty} \left(1+\frac{b_n}{b_{n+1}}\right)$
$=1+\lim\limits_{n \to \infty} \frac{b_n}{b_{n+1}}$
$=1+\frac{1}{\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}}$
$=1+\frac{1}{\mathcal{L}}$
Now the value of $\mathcal{L}$ can be easily obtained
Final Step
i.e., $\mathcal{L}=1+\frac{1}{\mathcal{L}}$
$\Rightarrow \mathcal{L}^2-\mathcal{L}-1=0$
$\Rightarrow \mathcal{L}=\frac{1\pm \sqrt{5}}{2}$
$\Rightarrow \mathcal{L}=\frac{1+\sqrt{5}}{2}$ [Since $a_n>0$] [ANS]
Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence).
Let $a_n=\frac{b_{n+1}}{b_n}$ where $b_1=1, b_2=1$ and $b_{n+2}=b_n+b_{n+1}$ , Then $\lim\limits_{n \to \infty} a_n$ is
Real Analysis
Sequence of Reals
Limit of a Sequence
But try the problem first...
Answer: $\frac{1+\sqrt5}{2}$
IIT JAM 2018 (Problem 2)
Advanced Calculus by Patrick Fitzpatrick
First hint
Given that, $a_n=\frac{b_{n+1}}{b_n}$
$\Rightarrow \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \frac{b_{n+1}}{b_n}= \mathcal{L} $ (say)
Now we know that , $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} a_{n+1} $
$\Rightarrow \mathcal{L}=\lim\limits_{n \to \infty} a_{n+1}$
Can you find an equation on $\mathcal{L}$ from which the value of $\mathcal{L}$ can be obtained.
Second Hint
$\mathcal{L}= \lim\limits_{n \to \infty } a_{n+1}$
$= \lim\limits_{n \to \infty} \frac{b_{n+2}}{b_{n+2}}$
$=\lim\limits_{n\to \infty} \frac{b_{n+1}+b_n}{b_{n+1}}$ [By the given recurrence relation]
$=\lim\limits_{n\to \infty} \left(1+\frac{b_n}{b_{n+1}}\right)$
$=1+\lim\limits_{n \to \infty} \frac{b_n}{b_{n+1}}$
$=1+\frac{1}{\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}}$
$=1+\frac{1}{\mathcal{L}}$
Now the value of $\mathcal{L}$ can be easily obtained
Final Step
i.e., $\mathcal{L}=1+\frac{1}{\mathcal{L}}$
$\Rightarrow \mathcal{L}^2-\mathcal{L}-1=0$
$\Rightarrow \mathcal{L}=\frac{1\pm \sqrt{5}}{2}$
$\Rightarrow \mathcal{L}=\frac{1+\sqrt{5}}{2}$ [Since $a_n>0$] [ANS]
