The problem is, let Z be a standard normal variable and $n\geq1$ be an integer. Show that $E[Z^{n+1}]=nE[Z^{n-1}]$. Here's what I've got so far, miraculously:
$E[Z^{n+1}]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{n+1}e^{-z^{2}/2}dz $
Doing integration by parts with $u=z^n$ and $dv=ze^{-z^{2}/2}$ gives
$[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty+n\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-z^{2}/2}z^{n-1}dz=[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty +nE[Z^{n-1}]$
Now my Calc II teacher actually told me never to do this $[f(x)]_{-\infty}^\infty$ but my probability textbook does it so I'm just gonna throw caution to the wind. I'm thinking I turn $[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty$ into $\frac{1}{2\pi}(\lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}-\lim_{z\rightarrow -\infty}-z^{n}e^{-z^{2}/2})$Wolfram Alpha tells me that each of those limits is 0 but I don't quite see how—as far as I can see both of those limits are in indeterminate forms. What I was thinking is maybe since $n$ is an integer greater than 1 that I can argue that if I write $lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}$ as $lim_{z\rightarrow\infty}-z^n/e^{z^2 /2}$ then repeated applications of l'Hopital's rule will eventually put a 0 in the numerator while you'll still have some exponential function in the denominator, and so that'll be zero and hence the limit is zero. But that feels cumbersome and I'm not sure what I can do beyond just asserting that, unless I want, like, prove that by induction on n or something. Am I missing some obvious better way to evaluate this limit?