Question

Converge or Diverge?

Answer 1

What are all the values of p for which \[\int\limits_{1}^{\infty} \frac{ 1 }{ x ^{2p} }dx\] converges?

a) p < -1
b) b > 0
c) p > .5
d) p > 1
e) there are no values of p for which this integral converges.



My problem is that im getting confused on the difference between converge and diverge,

Answer 2

@zepdrix

Answer 3

$$ \Large {
\int_{1}^{\infty } \frac{1}{x^{2p}} = \lim _{x \to \infty} \frac{x^{-2p+1}-1}{-2p+1}


}$$

Answer 4

this converges for -2p+1 <0 , otherwise x is not in the denominator

Answer 5

$$
\Large {
\int_{1}^{\infty } \frac{1}{x^{2p}} = \begin{cases}
\lim _{x \to \infty} \frac{x^{-2p+1}-1}{-2p+1} ~\rm ~~if~ p \neq 0.5
\\ \lim _{x \to \infty} \ln x \rm ~~~~if ~ p = 0.5
\end{cases}

}

$$

Answer 6

this converges for -2p+1 <0 , otherwise x has a positive exponent and will diverge as x goes to infinity.
if you solve the inequality -2p + 1 < 0 , you should get your answer