8. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=\frac{1}{4}, a_{n+1}=a_{n}+a_{n}^{2}\left(n \in \mathbf{Z}_{+}\right) . \\
\text {Let } T_{2020}=\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{2020}+1} .
\end{array}
$$

If the value of $T_{2020}$ lies in the interval $(k, k+1)$, then the integer $k$ is $\qquad$ .
8.3.

From the problem, we know
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow T_{2020}=\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}{a_{3}}+\cdots+\frac{1}{a_{2020}}-\frac{1}{a_{2021}} \\
\quad=\frac{1}{a_{1}}-\frac{1}{a_{2021}}=4-\frac{1}{a_{2021}} .
\end{array}
$$

To prove $T_{2020} \in(3,4)$, we only need to prove $a_{2021}>1$.
From $a_{n+1}=a_{n}+a_{n}^{2}>a_{n}$
$$
\begin{aligned}
\Rightarrow & a_{17}=a_{16}^{2}+a_{15}^{2}+\cdots+a_{1}^{2}+a_{1} \\
& >16 a_{1}^{2}=1 .
\end{aligned}
$$

By induction, we know for any $n>17, a_{n}>1$, then $T_{2020} \in(3,4)$, i.e., $k=3$.