Question

(Calculus) Need help with Root Test!

Answer 1

Here is my problem

Answer 2

I understand how to take the "nth" root of this and use that to get convergence or divergence, I just don't understand how to turn it into a form of e^x to use L'hopital's rule.

Answer 3

You sure you need to use L'Hopitals?

Answer 4

I feel like it could be accomplished without it, but I think my professor wants me to know how to do it with L'Hopitals for my upcoming test..

Answer 5

Well after taking the nth root we end up with \(\Large \frac{7n^{1/n}-1}{4n^{1/n} -1} \) right?

Answer 6

Yessir, I cant' figure out how to turn the n^(1/n) into an e^x form function.

Answer 7

would it turn into e^(ln(7n)) -1 and e^(ln(4n)-1? I,m thinking that's right.

Answer 8

I know what your professor is talking about.. but I don't remember how to get there. Maybe @saifoo.khan knows

Answer 9

Alright, I sent him a message.

Answer 10

Sorry, I forgot L'Hopital. :/

Answer 11

Is there any help you could give in terms of expressing the function into terms of e^x?

Answer 12

Okay, I found some old notes.. good thing I kept them. But from I see, \(\Large (1+\frac{1}{n})^n = e \)

Answer 13

So, it would turn into ((e^7)-1)/((e^4)-1)?

Answer 14

when the limit goes to infinity that is.. but we have to convert it somehow.. Amistre knows probably

Answer 15

The only example I have to go on is this: \[\sum_{0}^{\infty} (1+\frac{1}{n})^{n^2} \]

Which issn't much :/

Answer 16

hmm

Answer 17

You'd take the nth root, that'll leave with what's inside, plus the n power, which is e when it goes to \(\infty \)

Answer 18

Yeah, i see that.

Answer 19

When I put this in SymboLab I get a divergent series as the limit goes to infinity as well.

Answer 20

Hey, I'll be back later. I have to go do something..

Answer 21

Alright, sorry I couldn't help much :/

Answer 22

Well, unless rational can figure something out?

Answer 23

No worries, man. You tried :)

Answer 24

still working on this ?

Answer 25

we can rewrite the general term of your series as follows:
\[{a_n} = \exp \left[ {n\ln \left( {\frac{{7\sqrt[n]{n} - 1}}{{4\sqrt[n]{n} - 1}}} \right)} \right]\]

Answer 26

so we get:
\[\sqrt[n]{{{a_n}}} = \exp \left[ {\ln \left( {\frac{{7\sqrt[n]{n} - 1}}{{4\sqrt[n]{n} - 1}}} \right)} \right]\]

Answer 27

then applying the n-th root criterion, we can write:
\[{\lim _{n \to \infty }}\sqrt[n]{{{a_n}}} = \exp \left[ {{{\lim }_{n \to \infty }}\ln \left( {\frac{{7\sqrt[n]{n} - 1}}{{4\sqrt[n]{n} - 1}}} \right)} \right]\]

Answer 28

now, keeping in mind that:
\[{\lim _{n \to \infty }}\sqrt[n]{n} = 1\]

Answer 29

we have:
\[{\lim _{n \to \infty }}\sqrt[n]{{{a_n}}} = \exp \left[ {{{\lim }_{n \to \infty }}\ln \left( {\frac{{7\sqrt[n]{n} - 1}}{{4\sqrt[n]{n} - 1}}} \right)} \right] = \exp \left( 2 \right) > 1\]

Answer 30

In other words, your series is a divergent series

Answer 31

Thanks a bunch!