CLT for random variables with varying distributions The following is not a homework problem. Here it is:

Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random variables such that
$$
P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$

$$
P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}.
$$
Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.

Now, I am guessing that the given probabilities are for $k\geq 2$, since $k=1$ would yield an inconsistent definition. Furthermore, I am assuming the case $k=1$ corresponds to only the second line.

The first sign of trouble comes from the fact that these random variables are not identically distributed (otherwise, we could just use CLT). One thing I noticed is that $\mathbb{E}(X_k)=0$ for each $k$. Also, it appears as though the limit of the sequence is a random variable $X$ with $P(X=1)=P(X=-1)=\frac{1}{2} \hspace{2mm}$  (not that this result is necessarily of much help).

I'd very much appreciate any suggestions on how to approach this problem. Thanks!
Let $Y_k=\mathrm{sgn}(X_k)$, then $(Y_k)$ is a centered i.i.d. Bernoulli sequence hence, by the most usual CLT, $T_n=\frac1{\sqrt{n}}\sum\limits_{k=1}^nY_k$ converges in distribution to a standard normal random variable $T$. By Borel-Cantelli lemma, the random set $\{n\geqslant1\mid X_n\ne Y_n\}$ is almost surely finite hence there exists some almost surely finite random variable $Z$ such that $\left|\sum\limits_{k=1}^nX_k-\sum\limits_{k=1}^nY_k\right|\leqslant Z$ for every $n$. In particular $R_n-T_n\to0$ almost surely, where $R_n=\frac1{\sqrt{n}}\sum\limits_{k=1}^nX_k$.

Now it is a general fact that if $T_n\to T$ in distribution and if $R_n-T_n\to0$ almost surely then $R_n\to T$ in distribution. Thus, CLT holds for $(X_n)$ as well.
Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random
variables such that $$ P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$

$$ P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}. $$ Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
The characteristic function of $X_k$ is
$$\phi_k(\xi)=(\frac{1}{2}-\frac{1}{2k^2})e^{i\xi}+(\frac{1}{2}-\frac{1}{2k^2})e^{-i\xi}+\frac{1}{2k^2}e^{ik\xi}+\frac{1}{2k^2}e^{-ik\xi}$$
$$=(1-\frac{1}{k^2})\cos \xi +\frac{1}{k^2}\cos k\xi = \cos \xi +\frac{1}{k^2}(\cos k\xi-\cos\xi)$$
Then the characteristic function of $X_k/\sqrt{n}$ is
$$Ee^{i\xi X_k/\sqrt{n}}=\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})$$
and
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}}=\prod_{k=1}^n\left(\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})\right)$$
$$=\exp\left[\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)\right]$$
Then
$$\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}}-\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}})+\sum_{k=1}^n\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=S_1+S_2$$
For $S_1$ we have
$$S_1=-n(1-\cos\frac{\xi}{\sqrt{n}})=-2n
\sin^2\frac{\xi}{\sqrt{n}}=-2n\frac{\sin^2\frac{\xi}{\sqrt{n}}}{\frac{\xi^2}{n}}\times\frac{\xi^2}{n}
\to -2\xi^2 , \ \ n \to \infty.$$
$$S_2=\sum_{k=1}^n\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=-2\sum_{k=1}^n\frac{1}{k^2}\sin\frac{(k+1)\xi}{2\sqrt{n}}\sin\frac{(k-1)\xi}{2\sqrt{n}}$$
$$=-2\sum_{k=1}^n\frac{1}{k^2}\frac{(k^2-1)\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=-2\sum_{k=1}^n\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }+
2\sum_{k=1}^n\frac{1}{k^2}\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=S_{21}+S_{22}.$$

Now it is clear that $S_{21} \to -2\xi^2$ and $S_{22} \to 0$ as $n \to \infty$.
Therefore
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}} \to e^{-4\xi^2}, \ \ n \to \infty.$$
This is the characteristic function of r.v. $X=2Z \sim N(0,4)$.