TIFR CSE 2013 | Part A | Question: 19

Question

Consider a sequence of numbers $\large (\epsilon _{n}: n= 1, 2,...)$, such that $\epsilon _{1}=10$ and

$\large \epsilon _{n+1}=\dfrac{20\epsilon _{n}}{20+\epsilon _{n}}$

for $n\geq 1$. Which of the following statements is true?

Hint: Consider the sequence of reciprocals.

• A. The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ converges to zero.
• B. $\large \epsilon _{n}\geq 1$ for all $n$
• C. The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ is decreasing and converges to $1.$
• D. The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ is decreasing and then increasing. Finally it converges to $1.$
• E. None of the above.

Answer 1

$\large\epsilon_1$ is positive.

In the formula for $\large\epsilon_{n+1}$, we only add, multiply and divide positive numbers. Thus, all $\large \epsilon_n$ are positive.

Also, $\large\epsilon_{n+1} <  \epsilon_n$

Proof:

$\large\begin{align}
\epsilon_{n+1} - \epsilon_n &= \frac{20 \cdot \epsilon_n}{20+\epsilon_n} -  \color{red}{\epsilon_n}\\[1em]
&= \frac{20 \cdot \epsilon_n  \color{red}{-20\cdot \epsilon_n - (\epsilon_n)^2}}{20+\epsilon_n}\\[1em]
&= \frac{{\Large\color{red}-}(\epsilon_n)^2}{20+\epsilon_n}\\[1em]
&< 0\\[1em]
\hline
\epsilon_{n+1} - \epsilon_n &< 0\\[1em]
\epsilon_{n+1} &< \epsilon_n
\end{align}$

Thus, the sequence is decreasing.

Since the sequence is decreasing and is bounded below by $0$, we know that the sequence converges (Monotone Convergence Theorem).

The only fixed point of the sequence can be found as follows:

$\large\epsilon_f = \dfrac{20 \cdot \epsilon_f}{20+\epsilon_f}$

$\large20 \cdot \epsilon_f + (\epsilon_f)^2 = 20 \cdot \epsilon_f$

$\large(\epsilon_f)^2= 0$

$\large \epsilon_f = 0$

Hence, the sequence converges to $0$.

Option a is correct.

Comments

Are you kidding me?

 

$x_1=10$ (given)

$x_2=\frac{20}3$

$x_3=5$

$x_4=4$

$x_5=\frac{10}3$

$x_6=\frac{20}7$

 

How on Earth did you find any $x_n$ to be 20?

---

@Pragy Agarwal 

Extremely sorry , Focusing on "n" , i forgot Xn+1.

Its my bad 😔

Thank you for giving your precious time 🙂

---

@Pragy Agarwal Sir, 

any source wheré i can find how to solve where series converges.

Answer 2

A hint is given that consider the reciprocal. So, doing that will leave you with the general formula as 20/(n + 2) for (n + 1)th term. Now, it is clearly evident that this converges to zero as n increases.

Answer 3

Given, $\epsilon_{n+1} =\dfrac{20\epsilon_n}{20 + \epsilon_n}$; $\epsilon_1 = 10$

$$\begin{align}
\epsilon_{n+1} &= \dfrac{20\epsilon_n}{20 + \epsilon_n} = \dfrac{1}{\frac{1}{20} + \frac{1}{\epsilon_n}} \\
&= \dfrac{1}{\frac{1}{20} + \dfrac{1}{\frac{1}{\frac{1}{20} + \frac{1}{\epsilon_{n-1}}}}} = \dfrac{1}{\frac{1}{20} + \frac{1}{20} + \frac{1}{\epsilon_{n-1}}} = \dfrac{1}{\frac{2}{20} + \frac{1}{\epsilon_{n-1}}} \\
&= \dfrac{1}{\frac{3}{20} + \frac{1}{\epsilon_{n-2}}} \\
&= \qquad \vdots \\
&= \dfrac{1}{\frac{n}{20} + \epsilon_{1}} = \dfrac{1}{\frac{n}{20}+10}
\end{align}$$

$\lim_{n \to \infty} \ \epsilon_{n+1} = \lim_{n \to \infty} \ \dfrac{20}{n+200} = 0$

$\textbf{Option (A) is correct}$