Proving $M_n := e^{S_n - n/2}$ does not converge in $L^2$ Let $S_n = \sum_{i=1}^nX_i$, where $X_i \sim N(0,1)$ i.i.d. Let $M_n := e^{S_n - n/2}$ and $\mathcal{F}_n = \sigma(X_1, \dotsc, X_n)$, then $M_n$ is a martingale w.r.t. $\mathcal{F}_n$.
I need to show that $M_n$ does not converge in $L^2$.
I see that
\begin{align*}
\mathbb{E}[M_n^2] &= \mathbb{E}\left[e^{2S_n - n}\right]\\
&= e^{-n}\mathbb{E}\left[e^{2\sum_{i=1}^nX_i}\right]\\
&= e^{-n}\mathbb{E}\left[\prod_{i=1}^ne^{2X_i}\right]\\
&= e^{-n}\prod_{i=1}^n\mathbb{E}[e^{2X_i}] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Independent)}\\
&= e^{-n}\prod_{i=1}^ne^{2} = e^{-n}e^{2n} = e^n.
\end{align*}
The expectation squared of $M_n$ will grow exponentially as $n \to \infty$ and so I can conclude that $M_n$ does not converge in $L^2$. Is that correct?
 A: MSE is nagging me about extended discussions in comments, so I will post as an answer. I'll try to argue this avoiding general ideas about normed vector spaces.
First show that:
$$\sqrt {\mathrm E[(X + Y)^2]} \le \sqrt {\mathrm E[X^2]} + \sqrt {\mathrm E[Y^2]}$$
You'll need to know the Cauchy-Schwartz equality:
$$\mathrm E[|XY|] \le \sqrt{\mathrm E[X^2]} \sqrt {\mathrm E[Y^2]}$$
Then expanding:
$$\begin{align*}\mathrm E[(X + Y)^2] & = \mathrm E[X^2] + 2 \mathrm E[XY] + \mathrm E[Y^2] \\ & \le \mathrm E[X^2] + 2 \mathrm E[|XY|] + \mathrm E[Y^2] \\ & \le \mathrm E[X^2] + 2 \sqrt{\mathrm E[X^2]} \sqrt {\mathrm E[Y^2]} + \mathrm E[Y^2] \\ & = (\sqrt {\mathrm E[X^2]} + \sqrt {\mathrm E[Y^2]})^2\end{align*}$$
Then we get the desired inequality by taking square roots. You can rearrange this to (mimic the proof of the reverse triangle inequality):
$$\left|\sqrt {\mathrm E[X_n^2]} - \sqrt {\mathrm E[X_\infty^2]}\right| \le \sqrt {\mathrm E[{(X_n - X_\infty)^2]}}$$
If we had $X_n \to X_\infty$, we would have $\sqrt {\mathrm E[{(X_n - X_\infty)^2]}} \to 0$, so $\sqrt {\mathrm E[X_n^2]} \to \sqrt {\mathrm E[X_\infty^2]}$. But $\sqrt {\mathrm E[X_n^2]} \to \infty$, so $\sqrt {\mathrm E[X_\infty^2]} = \infty$, so $X_\infty$ can't be in $L^2$, contradiction.
We call $\lVert X\rVert_2 = \sqrt {\mathrm E[X^2]}$ the $L^2$ norm, and it makes $L^2$ into a "normed vector space".
