Question

an = [(ln(n))^7]/sqrt(n). Find the limit if it converges, indicate divergence if it diverges. How do I solve this?

Answer 1

Consider the end behavior of the continuous function\[f(x)=\frac{(\ln(x))^7}{\sqrt{x}}\]Numerator and denominator are both divergent to positive infinity, so use L'Hopital's rule.

Answer 2

After applying L'Hopital's rule, I got

\[(7(\ln x)^62\sqrt{x})/x\]

Am I deriving incorrectly?

Answer 3

Pretty close to what I got, but I didn't simplify the denominator. If you move the 1/x term to the denominator, your answer looks like\[f_1(x)=\frac{7(\ln(x))^6}{\sqrt{x}}\]Now repeat L'Hopital's rule a few more times, and you will eventually get a constant divided by the square root of x. I think that probably converges, as does your sequence.

Answer 4

after you apply L'hospital rule 7 times you will get

\( \large \frac{7*6*5*4*3*2*1}{(1/2)^7sqrt(x)}\)

Answer 5

OOOPS!! Dropped a factor in the derivative of the denominator. Hate it when that happens. Regardless, it doesn't change the end behavior of the continuous function, or the convergence of the sequence.

Answer 6

aqua1995, you've got this, right?

Answer 7

I'm still confused... I thought you had to take the derivative of the denominator was well when applying L'Hospital's Rule. Wouldn't that give you (1/2)n^(-1/2) in the denominator?

Answer 8

Yeah, it would, but there is a factor of 1/x in the numerator. If we move that term to the denominator, you get sqrt(x) there. We can move the factor of 1/2 to the numerator as two, also (that is the factor that I goofed on earlier). So after seven repetitions of L'Hopital's rule, we get \[f_7(x)=\frac{2^7*7!}{\sqrt{x}}\]which converges.

Answer 9

See it now?

Answer 10

I see where you're going with the 2, but would the denominator be 1/sqrt(x) and not just sqrt(x)?

Answer 11

No. When we take 1/x in the numerator, the one (factor) stays in the numerator, and the x moves to the denominator. With the existing factor of 1/sqrt(x), it becomes sqrt(x).

Answer 12

Hope I was helpful. I've gotta go to class now.

Answer 13

Ahh, I think I got it. Thanks so much! :-)

Answer 14

So this means it converges to zero?

Answer 15

Yup.