Let $(X_n)$ be an independent sequence of random variables such that $P(X_n = 1) = 1/2n, P(X_n = -1) = 1/2n, P(X_n = 0) = 1-1/n$
Set $Y_1 = X_1, Y_n = X_n1_{(Y_{n-1} = 0) } + nY_{n-1}\lvert X_n \rvert 1_{(Y_{n-1} \neq 0) }$.
I have already established that $(Y_n)$ is a martingale.
So we have $E[Y_{n+1} \mid Y_1, ... Y_n] = Y_n \implies E[E[Y_{n+1} \mid Y_1, ... Y_n]] = E[Y_n]$
thus $E[Y_{n=1}] = E[Y_n] = .... = E[Y_1] = E[X_1] = 1\frac{1}{2} - 1\frac{1}{2} + 0(1-1) = 0 $
It seems that it does not converge though because:
$E[\lvert Y_{n+1} \rvert] = E[E[\lvert Y_{n+1} \rvert \mid Y_1,...,Y_n]] = E[\frac{1}{n}1_{(Y_n = 0) }+\lvert nY_n\lvert X_n\rvert\rvert1_{(Y_n \neq 0)} \mid Y_1, ... Y_n] \\ = \frac{1}{n}P(Y_n = 0 \mid Y_1,...,Y_n) + n|Y_n|E[\lvert X_n \vert] = \frac{1}{n}P(Y_n = 0 \mid Y_1,...,Y_n) + n|Y_n|\frac{1}{n}$
But then I have to consider what $\lvert Y_n \rvert$ is and so on but seems it could take on increasingly larger values as $n \to \infty$
Note that if $X_n=0$, then $Y_{n+1}=0$.
By Borel-Cantelli, “$X_{n^2}=0$ for large enough $n$“ has probability $1$, so almost surely $Y_n$ is zero infinitely many times.
Thus, when $Y_n$ converges, its limit is (except an event of null probability) zero. But $Y_n$ is always a sequence of integers, so that in this case, $Y_n=0$ for all large enough $n$.
But it’s easy to see that $\{Y_n=0 \text{ for all large enough }n\} \subset \{X_n=0 \text{ for all large enough } n\}$, and the latter set has probability zero, by Borel-Cantelli again. So “$Y_n$ converges” has probability zero.