Determine if a sequence of independent rv's satisfies the Lindeberg condition 
Determine if the following sequences $\left\{A_k\right\}$ of independent r.vs satisfy the Lindeberg condition:
  
  
*
  
*$(a.)$ $A_k \sim \text{Poisson}\left(2^{-k}\right)$
  
*$(b.)$ $P(A_k = 1) =  P(A_k = -1) = \dfrac{1-2^{-k}}{2};\quad 
    P(A_k = 2^k) = P(A_k = -2^k) = \dfrac{1}{2^{k+1}}$
  

My solution: 


*

*For part $(a.)$: we would show that sequence $A_1, A_2,\ldots$ are NOT uniformly asymptotically negligible, so it does not satisfy Lindeberg condition. Note that since $A_k\sim \text{Poisson}\left(2^{-k}\right), $ $\sigma_{k}^2 = 2^{-k}$. This implies $\max_{k\leq n} \sigma_k^2 = 2^{-1}$. In addition $$\sum_{k=1}^{n} \sigma_k^2 = \sum_{k=1}^{n} 2^{-k} \approx \int_{1}^{n} \frac{1}{2^k} = \frac{-2^{-n}}{\ln2}$$ Therefore, $\displaystyle\lim_{n\rightarrow \infty} \dfrac{\max_{k\leq n} \sigma_k^2}{\sum_{k=1}^{n} \sigma_k^2} = \lim_{n\rightarrow \infty} -2^{n}\ln(2) = \infty > 0$. Thus we're done.

*For part $(b.)$: due to the given formulas of the discrete probability of $A_k$, $E(A_k) = 0$ for all $k=1,2,\ldots$ Thus 
$$E(A_k^2) = \sigma_k^2=\begin{cases}1,&k=1\\1-2^{-k}+2^k, &k\ge2\end{cases}$$ Thus, $\sum_{k=1}^{n} \sigma_k^2 = 1 +\ldots + (1-2^{-n}+2^n)$ and $\max_{k\leq n} \sigma_{k}^2 = 1 + 2^n - 2^{-n}.$ This implies $\displaystyle \lim_{n\rightarrow \infty} \dfrac{\max_{k\leq n} \sigma_{k}^2}{\sum_{i=1}^{n} \sigma_i^2} = 0$ by dividing both numerator and denominator by $1+2^n - 2^{-n}$, and note that this function is strictly increasing in $n$. Thus, the sequence $A_1, A_2,\ldots$ is uniformly asymptotically negligible. Finally, in order to apply Linderberg-Feller's theorem, it's sufficient to show $\dfrac{1-2^{-n}+2^{n}}{(1 +\ldots + 1-2^{-n}+2^n)^{1/2}}$ converges to $N(0,1)$ in distribution. But I'm getting stuck here.
My question: Could someone please help complete my proof above? Otherwise, any inputs on my proof above would also be appreciated.
 A: *

*For $(a.)$: $\sum_{k=1}^nσ^2_k$ is a geometric sum, so $$\sum_{k=1}^n 2^{-k}=-1+\sum_{k=0}^n2^{-k}=-1+\frac{1-2^{-n-1}}{1-\frac12}=1-2^{-n}$$ Since, $σ^2_k=2^{-k}$ is a decreasing sequence, you have that $$\lim_{n\to\infty}\frac{\max_{1\le k\le n}σ^2_k}{\sum_{k=1}^nσ^2_k}=\lim_{n\to\infty}\frac{σ^2_1}{1-2^{-n}}=\frac{\frac12}{1-0}=\frac12\neq 0$$

*For $(b.)$: I do not agree with $σ^2_1=1$. I think that $σ^2_k=1-2^{-k}+2^k$ for any $k\ge 1$ (but anyway this does not affect the solution). Again, using the geometric sum
$$\sum_{k=1}^nσ_k^2=\sum_{k=1}^n(1+2^k)-\sum_{k=1}^n2^{-k}=n+\frac{1-2^{n+1}}{1-2}-\frac{1-2^{-n-1}}{1-\frac12}=n+2^{n+1}-3+2^{-n}$$ Hence, since $σ^2_k$ is increasing, you have that $$\lim_{n\to\infty}\frac{\max_{1\le k\le n}σ^2_k}{\sum_{k=1}^nσ^2_k}=\lim_{n\to\infty}\frac{2^n+1-2^{-n}}{n+2^{n+1}+1-2^{-n}}=\lim_{n\to\infty}\frac{1+2^{-n}-2^{-2n}}{\frac{n}{2^{n}}+2+2^{-n}-2^{-2n}}=\frac12\neq 0$$ So, if I did not miscalculate, neither sequence of rv's satisfies the condition of asymptotically negligible variances.

