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How to determine if $\int_2^\infty x^2/e^x \; dx$ converges without computing it? I'm thinking of applying a comparison test but I'm not sure to what.

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Note that $\frac{x^2}{e^{x}} < \frac1{x^2}$ for $x>9$. Hence, split the integral from $2$ to $9$ and then from $9$ to $\infty$ and argue why both are finite.

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  • $\begingroup$ How did you obtain 9? $\endgroup$ – user93200 Mar 11 '12 at 19:43
  • $\begingroup$ @user93200: You know that $e^x$ grows faster than $x^4$, so you just have to find a point at which $e^x>x^4$, or $x>4\ln x$; from that point on you’ll have the desired inequality. A little work with a calculator quickly establishes $9$ as the smallest integer that works. $\endgroup$ – Brian M. Scott Mar 11 '12 at 19:55
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For every $x\geqslant2$, $x^2\mathrm e^{-x}\leqslant16\mathrm e^{-2}\mathrm e^{-x/2}$ hence $$\int\limits_2^{+\infty}x^2\mathrm e^{-x}\mathrm dx\leqslant16\mathrm e^{-2}\int\limits_2^{+\infty}\mathrm e^{-x/2}\mathrm dx=16\mathrm e^{-2}\cdot2\mathrm e^{-1}\lt2. $$

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Try comparing to $e^{-tx}$ where $0 < t < 1$.

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Look at the power series expansion of $e^x$. The $x^4$ term is $\dfrac{x^4}{4!}$, so for positive $x$, we have $e^x>\frac{x^4}{4!}$ and therefore $$\frac{x^2}{e^x} <\frac{4!}{x^2}.$$ We know that $$\int_2^\infty \frac{dx}{x^2}$$ converges, and the $4!$ on top makes no difference. Note that the same idea can be used mechanically to show, for example, that $\displaystyle\int_2^\infty \frac{x^{2012}}{e^x}dx$ converges, since for positive $x$, $e^x>\frac{x^{2014}}{2014!}$.

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  • $\begingroup$ In the 1st inequality, how do you obtain that x^2/e^x < than x^2/(only one term of the expansion of e^x)? I think because additional terms in the denominator can only make the ratio smaller. Why does the 4! on top makes no difference? Good answer, thank you. $\endgroup$ – user93200 Mar 11 '12 at 20:48
  • $\begingroup$ @user93200: Yes, for positive $x$ all the other terms are positive, so $e^x>x^4/4!$, and therefore $x^2/e^x<x^2/(x^4/4!)$, which is $4!/x^2$. As to $4!$ making no difference, for any non-zero constant $k$, $\int_2^\infty f(x)\,dx$ converges iff $\int_2^\infty kf(x)\,dx$ converges. $\endgroup$ – André Nicolas Mar 11 '12 at 21:00

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