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I was testing the convergence of the integral $$\int_0^{+\infty}\frac{\sqrt{x+1}}{1+2\sqrt x+x^2}dx$$ by calculating the limit $$\lim_{x\to+\infty}\frac{\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}}{\frac{1}{x^{3/2}}}=1>0$$ and by noting that $$\int_0^{+\infty}\frac{1}{x^{3/2}}=+\infty$$ therefore the original integral has to be divergent as well. However Mathematica seems to evaluate the original integral to 2.04517. What is wrong here?

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The limit you have found $$ \lim_{x\to+\infty}\frac{\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}}{\frac{1}{x^{3/2}}}=1 $$ implies that, as $x \to \infty$, $$ \frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}\sim{\frac{1}{x^{3/2}}} $$

you can't deduce it holds as $x \to 0$. Therefore, we have that, for any large fixed $B>0$, $$ \int_B^\infty\frac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}\:dx, \quad\int_B^\infty {\frac{1}{x^{3/2}}}\:dx, $$ are both convergent. Observe that the integrand $\dfrac{\sqrt{x+1}}{1+2\sqrt{x}+x^2}$ is a continuous function as $x \to 0^+$, giving the convergence of the initial integral over $(0,\infty)$.

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  • $\begingroup$ Doesn't $B$ have to be $\ge 1$ to deduce convergence for both of them, since the integral we're comparing to converges only for $B\ge 1$ ? $\endgroup$ – ahra Jul 8 '17 at 10:01
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    $\begingroup$ @ahra Not really, the integral $\int_B^\infty {\frac{1}{x^{3/2}}}\:dx$ converges for any $B>0$. Thus $B\ge1$ is fine, but necessary. $\endgroup$ – Olivier Oloa Jul 8 '17 at 10:03
  • $\begingroup$ Thank you for your answer! $\endgroup$ – ahra Jul 8 '17 at 10:24

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