This is the Exercise 27.6 from Billingsley.
Suppose $Z_1, Z_2, \dots$ are i.i.d. r.v.'s with mean $0$ and variance $1$, and define $X_{nk} = \sigma_{nk}Z_k$. Show that if $$\frac{\max_{1\le k \le n} \sigma^2_{nk}}{s^2_n} \rightarrow 0,$$ then the Lindeberg's condition is satisfied.
Where $s^2_n = \sum_{k=1}^n \sigma^2_{nk}$. I have done some attempts, but I am not sure if all the steps are legit or not.
First attempt (additional conditions)
Suppose additionally that $\mathbb{E}(Z_k^4) <\infty$. Then, the Lyapounov condition with $\delta = 2$ should holds, which implies the Lindeberg's one. I did the following steps
$$\frac{1}{s^4_n}\sum_{k=1}^n \mathbb{E}(|X_{nk}|^4) = \mathbb{E}(Z_1^4) \frac{1}{s^4_n}\sum_{k=1}^n \sigma_{nk}^4 \le \frac{\max_{1 \le k \le n}\sigma^2_{nk}}{s^2_n} \cdot \sum_{k=1}^n \frac{\sigma^2_{nk}}{s^2_n} = \frac{\max_{1 \le k \le n}\sigma^2_{nk}}{s^2_n} \rightarrow 0.$$
Second attempt
Without the further assumption, we have
$$\frac{1}{s^2_n}\sum_{k=1}^n \mathbb{E}(|X_{nk}|^2 1_{|X_nk|> \epsilon s_n}) = \frac{1}{s^2_n}\sum_{k=1}^n \sigma^2_{nk} \mathbb{E}(Z_1^2 1_{|Z_1| > \epsilon s_n/\sigma_{nk}}) \\ \le \frac{\max_{1 \le k \le n}\sigma^2_{nk}}{s^2_n} \sum_{k=1}^n \mathbb{E}(Z_1^2 1_{|Z_1| > \epsilon s_n/\sigma_{nk}})$$
Now call $a_n = s_n/\sigma_{nk}$. It should hold that $a_n \rightarrow \infty$. Therefore $\{Z_1^2 1_{|Z_1| > \epsilon a_n} \} \downarrow \emptyset $ which implies that for $n$ sufficiently large we have (?) $$\mathbb{E}(Z_1^2 1_{|Z_1| > \epsilon a_n}) < 1,$$ which implies (?) that $$\sum_{k=1}^\infty \mathbb{E}(Z_1^2 1_{|Z_1| > \epsilon a_n}) < \infty.$$ So, the Lindeberg condition should be satisfied.
It is true that $\mathbb{E}(Z_1^2 1_{|Z_1| > \epsilon a_n}) < 1$, but this does not imply the convergence of the series you mentioned.
Use instead the bound $$\mathbb{E}\left[Z_1^2 1_{|Z_1| > \epsilon \frac{s_n}{\sigma_{nk}}}\right] \leqslant \mathbb E\left[[Z_1^2 1_{|Z_1| > \epsilon M_n^{-1/2} }\right]$$ valid for each $k\in \{1,\dots,n\}$, where $$M_n:=\frac{\max_{1\leqslant k \leqslant n} \sigma^2_{nk}}{s^2_n}.$$