$\large\epsilon_1$ 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.