The sequence $(x_n)$: $\left\{\begin{array} .x_1=a>1\\ 2021x_{n+1}=x_n^2+2020x_n \end{array}\right.$ The sequence $(x_n)$ is defined by the fomula:
$$\left\{\begin{array}
.x_1=a>1\\
2021x_{n+1}=x_n^2+2020x_n \quad (*)
\end{array}\right., \quad n=1,2,3\dots$$
Find $\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\frac{x_i}{x_{i+1}-1}$
I see $(*)\Rightarrow2021(x_{n+1}-1)=x_n^2+2020x_n-2021=(x_n-1)(x_n+2021)$
So, $\frac{x_n}{x_{n+1}-1}=\frac{2021x_n}{(x_n-1)(x_n+2021)}$.
I have a hunch it helps, but, I don't do more. Please help me. Thanks.

Notice
$x_{n+1} = \frac{x_n(x_n+2020)}{2021} = x_n + \frac{x_n(x_n-1)}{2021}$, we have
$$\begin{align}\frac{x_n}{x_{n+1}-1} 
&= \frac{x_n(x_n-1)}{(x_n-1)(x_{n+1}-1)}
= 2021\frac{x_{n+1}-x_n}{(x_n-1)(x_{n+1}-1)}\\
&= 2021\left(\frac{1}{x_n -1} - \frac{1}{x_{n+1}-1}\right)
\end{align}
$$
The sum at hand is a telescoping sum and
$$\sum_{k=1}^n \frac{x_k}{x_{k+1}-1} =
2021\left(\frac{1}{x_1-1} - \frac{1}{x_{n+1}-1}\right)$$
When $x_1 = a > 1$, it is easy to see $x_n$ is an increasing sequence.
Furthermore, $x_{n+1} - x_n = \frac{x_n(x_n-1)}{2021} \ge \frac{a(a-1)}{2021}$ implies
$$\lim_{n\to\infty} x_n = \infty \quad\implies\quad
\lim_{n\to\infty} \frac{1}{x_{n+1} - 1 } = 0$$
As a result,
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{x_k}{x_{k+1}-1}
= \frac{2021}{x_1 - 1} = \frac{2021}{a-1}$$
