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When does $\int_1^\infty \frac{x^a}{1+x^a} dx$ converge? It doesn't look like convergent because

$$\int_1^\infty \frac{x^a}{1+x^a} dx \ge \int_1^\infty \frac{1}{1+x^a}dx$$

and the right hand side reminds me of $\int_1^\infty \frac{1}{1+x}$.

I also looked at the sum $\sum_1^\infty \frac{n^a}{1+n^a}$. Ratio test yields

$$\frac{(n+1)^a}{1+(n+1)^a}\frac{n^a +1}{n^a}=\left(1+\frac{1}{n}\right)^a\frac{n^a + 1}{(n+1)^a + 1}$$

Not sure how to proceed from here?

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    $\begingroup$ I'm pretty sure that your ratio converges for $\alpha \ge 1$ and diverges for $0 \le \alpha < 1$. $\endgroup$ – steven gregory Aug 21 '16 at 2:31
  • $\begingroup$ The ratio and root tests never work on "rational-like" functions. The limit when all said and done is always 1. $\endgroup$ – David P Aug 21 '16 at 2:44
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If $\alpha > 0$ this clearly diverges, as the integrand does not approach zero.

$\dfrac{x^\alpha}{1+x^\alpha} = \dfrac{1}{x^{-\alpha} + 1}$

From here you can convince yourself that you need $-\alpha > 1$, i.e. $\alpha < -1$.

If you want to prove this in an elementary way, you may use (prove) the following lemma:

If $f,g$ are positive and continuous functions such that $\displaystyle \ \lim_{x\to\infty}\dfrac{f(x)}{g(x)}= 1$ then

$\displaystyle \ \int_1^{\infty} f(x) \ dx \ $ converges iff $\displaystyle \ \int_1^{\infty} g(x) \ dx\ $ converges.

For $\alpha \ge 0$ take $g(x)=1$. For $\alpha < 0$ take $g(x)=\dfrac{1}{x^{\alpha}}$

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    $\begingroup$ The integral can converge even if the integrand does not approach zero. Just saying ... $\endgroup$ – zhw. Aug 21 '16 at 4:02
  • $\begingroup$ That is true, but can only happen if the limit does not exist. $\endgroup$ – David P Aug 21 '16 at 4:20
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If you look at the ratio test with $u_n=\frac{n^a}{1+n^a}$ and develop the result as a Taylor series for large values of $n$, you will systematicall find that $$\frac{u_{n+1}}{u_n}\to 1$$ for any $a$ and then the ratio test is inconclusive.

But, for such a case, you can consider Raabe's test

$$R_a=\lim_{n\to\infty} n\left(\left|\frac{u_n}{u_{n+1}}\right|-1\right)$$ according to which the series will be absolutely convergent if $R_a>1$ and divergent if $R_a<1$. Trying for a few values of $a$, you should notice that $$R_{-1}=-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ $$R_{-11/10}=\frac{11}{10}+\frac{11}{200 n}+O\left(\frac{1}{n^{11/10}}\right)$$ $$R_{-12/10}=\frac{6}{5}+\frac{3}{25 n}+O\left(\frac{1}{n^{6/5}}\right)$$ Then, the series would converge if and only if $a<1$.

Going further, you could show that, in such a case $$I_a=\int_1^\infty \frac{x^a}{1+x^a} dx=\frac{H_{\frac{1}{2} \left(\frac{1}{a}-1\right)}-H_{\frac{1}{2 a}}}{2 a}$$ where appear generalized harmonic numbers. Effectively $I_a \to \infty$ when $a \to -1^{-}$.

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