Question regarding central limit theorem proof using Lyapunov's condition Let $X_n\sim \operatorname{independent} \operatorname{Uniform} (0,n^2)$. I need to find $a_n$ and $b_n$ such that $$\frac{\sum_{k=1}^n (X_k -a_n)}{b_n}$$ converges in distribution to a non-degenerate limit.
My apporach is to use Lindeberg condition (Lyapunov's condition)  to this and prove this converges to standard normal distribution.
This is my work so far:
$X_k= n^2 Z_k$ where $Z_k$ ~ $U[0,1]$ . 
Under  Lyapounov's condition i use $\delta =2$ . So $$E|X_k|^{2+\delta} = E|X_k|^4 = n^8E|Z_k|^4.$$
$$D_n =\sum_1^n \sigma^n_k = \sum_1^n \frac {n^4}{12}.$$
Using Lyapounov's condition $$\frac{E|X_k|^{4}}{D^2_n} = \frac{E|X_k|^4\sum_n^8}{(\sum\frac {n^4}{12})^2}.$$
if my work is correct , then this seems to be a divergent series so this as a results of that lindeberg condition will not satisfy. 
Can anyone help me figure out what did i do incorrectly ? 
Maybe there is an easy way to prove this rather than this way. 
 A: I don't think the Lyapunov condition is needed here. Try to verify the Lindeberg condition directly. Using your notation,
$$ D_n^2 = \sum\limits_{k=1}^n\frac{k^4}{12} = O(n^5) $$
Now, let $a_k = E[X_k] = \frac{k^2}{2}$. Then, $|X_k - a_k| \leq \frac{3}{2}k^2$. So, 
$$ \begin{eqnarray} \frac{1}{D_n^2}\sum\limits_{k=1}^nE[|X_k-a_k|^2\chi_{\{|X_k-a_k|>\epsilon D_n\}}]
&\leq& \frac{9}{4D_n^2}\sum\limits_{k=1}^nk^4P(|X_k-a_k|>\epsilon D_n)\\ 
&\leq& \frac{9}{48\epsilon^2D_n^4}\sum\limits_{k=1}^nk^8\end{eqnarray}$$
Since the sum behaves like $O(n^9)$ and the denominator behaves like $O(n^{10})$, the above quantity converges to $0$. So the Lindeberg condition holds. 
A: $\def\dto{\xrightarrow{\mathrm{d}}}$Your application of Lyapnov condition is indeed correct, but the resulting sequence is convergent. In fact, suppose $Z \sim U(0, 1)$ and for any $δ > 0$,$$
E(|X_k - E(X_k)|^{2 + δ}) = (k^2)^{2 + δ} E(|Z - E(Z)|^{2 + δ}) = c_1 k^{4 + 2δ}\\
\Longrightarrow \sum_{k = 1}^n E(|X_k - E(X_k)|^{2 + δ}) = c_1 \sum_{k = 1}^n k^{4 + 2δ} \sim c_2 n^{5 + 2δ}. \quad (n → ∞)
$$
Analogously,$$
s_n^2 = \sum_{k = 1}^n D(X_k) = \sum_{k = 1}^n \frac{k^4}{12} \sim c_3 n^5. \quad (n → ∞)
$$
Therefore,$$
\frac{1}{s_n^{2 + δ}} \sum_{k = 1}^n E(|X_k - E(X_k)|^{2 + δ}) \sim \frac{c_2 n^{5 + 2δ}}{(c_3 n^5)^{\frac{2 + δ}{2}}} = \frac{c_4}{n^{\frac{δ}{2}}} → 0, \quad (n → ∞)
$$
which by Lyaponov CLT implies$$
\frac{1}{s_n} \sum_{k = 1}^n (X_k - E(X_k)) \dto Y \sim N(0, 1).
$$
