$x_1=10$ (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?

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5 votes

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.

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

2 votes

Best answer

$\large\epsilon_1$ 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.**