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I need to check convergence of $\int_1^∞ \frac{1}{x^{\frac{1}{x}+1}} dx$ . I think it divergence cause it bigger than $\int_1^∞ \frac{1}{x} dx$ but I can't prove it. I have an hint that $\lim\limits_{x \to ∞} $$ x^{\frac{1}{x}}$ is 1 but I cant decipher how it useful. any help? THANKS

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  • $\begingroup$ Say we have an integral $\int_a^\infty f(x)\,\mathrm{d}x$ where $f(x)$ is strictly positive for $x \in [a,\infty]$. Can the integral converge if $\lim_{x \to \infty} f(x) \neq 0$? $\endgroup$ May 20, 2016 at 10:57
  • $\begingroup$ You mean that because $\lim\limits_{x \to ∞}f(x)=1 $ so the integral is necessarily divergent? Its a theorem? THANKS! $\endgroup$
    – A-H
    May 20, 2016 at 11:29

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Notice that $$\frac1{x^{\frac1x+1}}\sim_\infty \frac1x$$ which means that: for $\epsilon>0$ there is $A>0$ such that for $x\ge A$ we have $$\frac1{x^{\frac1x+1}}\ge (1-\epsilon)\frac1x$$

From this inequality we conclude that the given integral is divergent since $$\int_A^\infty\frac{dx}{x}=+\infty$$

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