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I want to use the comparison test - I thought of $1 \over x$. but this function is larger and also diverges so it doesn't help me much.

Thanks in advance!

user21312
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  • Your basic instincts are off. Use a calculator to calculate $x^{-x}$ for some values < 10 to get a correct understanding how your integrand behaves. Then take a look at the solution (better: hint) below. – Ingix Apr 27 '17 at 14:37
  • L'Hospital's rule does not show that it tends to $\infty$, since $x\ln x\to\infty$ (note that $\ln x/(1/x)$ is $\infty/0$, so you cannot use L'Hospital here). – Yuxiao Xie Apr 27 '17 at 14:37
  • You're right I edited easier than I thought Thank you for the clarification – user21312 Apr 27 '17 at 14:43

1 Answers1

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If $x>2$ hen $$ \frac{1}{x^x}<\frac{1}{x^2}. $$

Julián Aguirre
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