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Find the convergence region of the improper integral $$\int_1^\infty \frac{x^a}{1+x^a}dx\,.$$

To find the convergence region of this integral, it is equivalent to find the convergence region of the series $$\sum_{n=1}^\infty \frac{n^a}{1+n^a}\,.$$

I use the ratio test and root test but could find the convergence area for $a$.

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The integral is convergent for $-a \gt 1$ or $a\lt-1$ and otherwise it is divergent.

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  • $\begingroup$ I divide $n^a$ on both denominator and nominator. Then I get $\frac{1}{n^{-a}+1}$. Then how do you see that $-a>1$? $\endgroup$ – Kenneth.K Mar 25 '17 at 6:36
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    $\begingroup$ @Kenneth.K Note that $\int_{1}^{\infty}\frac {dx}{x^p}$ converges for $p\gt1$ $\endgroup$ – user409521 Mar 25 '17 at 6:50
  • $\begingroup$ I get $\frac{1}{n^{-a}}>\frac{1}{n^{-a}+1}>\frac{1}{(n+1)^{-a}}$. Therefore, $-a>1$ iff the integral is convergent? $\endgroup$ – Kenneth.K Mar 25 '17 at 7:00

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