How to show that $Y_n = \sum_{i=1}^n Z_i$ is a martingale. 
Let $Z_i$ be independent rvs such that
$$ P(Z_n = a_n) = \frac{1}{2n^2}, P(Z_n=0)=1-\frac{1}{n^2}, P(Z_n = -a_n) = \frac{1}{2n^2}$$
where $a_1=2, a_n = 4\sum_{j=1}^{n-1}a_j $ for $n\ge 2$.

*

*Show that $Y_n = \sum_{i=1}^n Z_i$ is a martingale.

*Show that $Y=\lim Y_n$ exists almost surely, but that there exists no $M$ such that $E|Y_n∣\le M for all $n$.



*

*$Z$ is discrete and it's distribution is on the probability space so it must be an adapted process, and $Y$ is the sum of adapted processes meaning it is also adapted.


*It is integrable as each $a_j$ is finite.
I'm not sure if these two make sense, are they OK?
To show convergence:
By Martingale Convergence show that $\lim_{n\to\infty} \mathbb E(Y_{n+1}-Y_n)^2=0$. So if we condition on $Y_t$, we get
\begin{align*}
    \mathbb E\left((Y_{t+1}-Y_t)^2|Y_t\right)&=a_n\left(\frac{1}{2n^2}-Y_n\right)^2 + (1-a_n)\left(\frac{1}{2n^2}-Y_n\right)^2+ 0\left(\frac{1}{n^2}-Y_n\right)^2\\
    &=\left(\frac{1}{2n^2}-Y_n\right)^2\\
    &\ge 0
\end{align*}
Hence, by the tower property for expectations, we can deduce that
$$ \mathbb E\left((Y_{t+1}-Y_t)^2\right) = \mathbb{E}\left(\left(\frac{1}{2n^2}-Y_n\right)^2\right) = \frac{1}{2n^2}- \mathbb{E}((Y_n)^2) $$
Therefore,
$\lim_{n\to\infty} \mathbb E(Y_{n+1}-Y_n)^2=- \mathbb{E}((Y_\infty)^2)$
 A: Perhaps the reference you are consulting has a different definition, but as I understand, to show that $Y_n$ is Martingale, you have to show that:
i) $Y_n \in \mathcal F_n , \quad \forall n\in \mathbb N$;
ii) $E|Y_n|<\infty ,\quad \forall n \in \mathbb N$;
iii)$ E[Y_n \mid \mathcal F_{n-1}] = Y_{n-1}$.
Now, as you pointed out. The two initial properties can be trivially shown. Only the last one remains. Hence,
$$E[Y_n \mid \mathcal F_{n-1} ]= E[Z_n + Y_{n-1} \mid \mathcal F_{n-1}]=
 E[Z_n] - Y_{n-1}= Y_{n-1}
$$
Note that above we used the fact that $Z_n$ is independent and that $E[Z_n] = \frac{an}{2n^2} - 0(1-1/n^2) -\frac{an}{2n^2} =0$.
So $Y_n$ is a  martingale.
Finally, you asked in the comments to prove that $Y_n$ converges almost surely. We can prove this using Borel-Cantelli as follows:
If for any $\epsilon>0$, we have $\sum^\infty_{n=1}P(|Y_n|>\epsilon)<\infty$, then, $P(\lim_{n\to\infty}Y_n = 0)=1$
So, for any $\epsilon>0$, we note that
$$P(|Y_n| > \epsilon) \leq P(|Y_n| >0) = P(Y_n = a_n) + P(Y_n = -a_n) = \frac{1}{n^2}$$
Therefore,
$$
\sum^\infty_{n=1}P(|Y_n|>\epsilon)<\infty\implies Y_n \rightarrow_{a.s}0
$$
This concludes the proof.
