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I know this can be done easily with just partial fractions but I'm wondering how I would show this converges using integral comparison test.

Question / Attempt:

$\int_{1}^{\infty} \frac{1}{x^2+x}dx$

For $x \in [1, \infty), f(x) = \frac{1}{x^2+x} \geq 0$

For $x \in [1, \infty), f(x) = \frac{1}{x^2+x} \leq \frac{1}{x^2} = g(x)$

Consider $$\lim_{A\to\infty}\int_{1}^{A} g(x) dx = \lim_{A\to\infty}\int_{1}^{A} \frac{1}{x^2} dx$$

$$= \lim_{A\to\infty} -\left(\frac{1}{x}\right)\bigg|_{1}^{A} = 1$$

Therefore $\int g(x)dx$ converges. Also therefore by the comparison test $\int f(x)dx$ converges as well.

Is this how you would would do comparison test for integral? Also whats the difference between the integral comparison test and the series comparison test in terms of hypothesis and methods?

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Observe that on given interval $x^2+x>x^2$ and so $\frac{1}{x^2+x}<\frac{1}{x^2}$. Therefore you can compare $\int_{1}^{\infty} \frac{1}{x^2+x}dx$with $\int_{1}^{\infty} \frac{1}{x^2}dx$, latter integral is $1$. On a side note, your given integral can be calculated even easier without partial fractions: Set $x=1/t$ and your integral changes into $\int_{0}^{1} \frac{1}{1+t}dt=ln2$.

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Yes $1/x^2$ would probably also be my first candidate for comparison. It seems correct to me. Series and (at least Riemann-)integrals are connected through Riemann sums which can give upper and lower bounds both ways. If you want to prove a series result you can use an integral if it seems easier. Or if you want to prove an integral result you can evaulate a series if it seems easier.

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  • $\begingroup$ Is there a difference in the hypothesis of the series comparison test? Do I just have to show that the function is positive and I can start the comparison for both? $\endgroup$ – user349557 Apr 16 '17 at 21:26
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The integrand $f$ is continue on $[1;+\infty[$. To prove that the integral is converging you can argue that $\lim_{x\rightarrow \infty} |f(x)|x^2=constant$

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