Use the integral test to determine whether the series is convergent or divergent. [infinity] n = 2 n2 n3 + 1 Evaluate the following integral. [infinity] 2 x2 x3 + 1 dx

Respuesta :

I think the given series is

[tex]\displaystyle\sum_{n=2}^\infty \frac{n^2}{n^3+1}[/tex]

You can use the integral test because the summand is clearly positive and decreasing. Then

[tex]\displaystyle\sum_{n=2}^\infty\frac{n^2}{n^3+1} > \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx[/tex]

Substitute u = x ³ + 1 and du = 3x ² dx, so the integral becomes

[tex]\displaystyle \int_2^\infty\frac{x^2}{x^3+1}\,\mathrm dx = \frac13\int_9^\infty\frac{\mathrm du}u = \frac13\ln(u)\bigg|_{u=9}^{u\to\infty}[/tex]

As u approaches infinity, we have ln(u) also approaching infinity (whereas 1/3 ln(9) is finite), so the integral and hence the sum diverges.

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