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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?

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Exponential function $f(x) = a^x$ of the base $a$ greater than 1 overwhelms any polynomial as $x \to \infty$. There are many ways to prove this fact, including the method using power series of $e^x$. That is, for $x > 0$, we have $$ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \geq \sum_{k=0}^{n} \frac{x^k}{k!}.$$ – Sangchul Lee Aug 7 '12 at 5:28
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The calculation using integration by parts is the important part. Now to the main question.

Your idea of using L'Hospital's Rule is good. Think of the limit of $|z|^n e^{-z^2/2}$ as $|z|\to \infty$. (This is just to collapse the two limit calculations into one.)

In order to make the repeated differentiation easier, one might as well let $u=z^2/2$. So $|z^n|=2^{n/2}u^{n/2}$. We are trying to find $$\lim_{u\to \infty}\frac{2^{n/2}u^{n/2}}{e^u}.$$ It is enough to show that for any positive integer $k$, $$\lim_{u\to\infty}\frac{u^k}{e^u}=0.$$ This is a routine application of L'Hospital's Rule.

Much more informally (but enough for probability), $e^{z^2/2}$ grows enormously faster in the long run than any power of $z$.

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Thanks, this is great! I had to go through carefully on paper to make sense of the u substitution but once I got it I got it. And yeah, the integration by parts was a wild stab in the dark—I mean there wasn't anything else to do with that expression but do integration by parts but I still didn't expect anything useful to come out of it and then, lo and behold, there's the result I want staring me in the face. Thanks for your help! – crf Aug 7 '12 at 16:33
Integration by parts can be used to find many Reduction Formulas of a character similar to the one you proved. A simpler example is expressing the integral of $\sin^{n}x$ (say over $0$ to $\pi/2$) in terms of the integral of $\sin^{n-2}x$. – André Nicolas Aug 7 '12 at 16:41

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