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Limit of a Sequence | IIT JAM 2018 | Problem 2

Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence).

Limit of a Sequence - IIT JAM 2018 (Problem 2)


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

  • $\frac{1-\sqrt5}{2}$
  • $\frac{1+\sqrt5}{2}$
  • $\frac{1+\sqrt3}{2}$
  • $\frac{1-\sqrt3}{2}$

Key Concepts


Real Analysis

Sequence of Reals

Limit of a Sequence

Check the Answer


Answer: $\frac{1+\sqrt5}{2}$

IIT JAM 2018 (Problem 2)

AdvancedĀ Calculus by Patrick Fitzpatrick

Try with Hints


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.

$\mathcal{L}= \lim\limits_{n \to \infty } a_{n+1}$

$= \lim\limits_{n \to \infty} \frac{b_{n+2}}{b_{n+2}}$

$=\lim\limits_{n\to \infty} \frac{b_{n+1}+b_n}{b_{n+1}}$ [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

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]

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