Question

Regarding limits at infinity when finding the sum of an infinite series, can you just apply l'hopital to the infinite limit when you get an indeterminate form, or do you have to have set it equal to a function, and take the limit of the function as x approaches infinity to be able to apply l'hopital's rule? The solutions manual I use keeps applying l'hopital's rule to the n's, and my understanding was that you either had to be able to solve the infinite limit or relate it to a function f(x) where you could take the limit as x approaches infinity and then apply l'hopitals to that function

Answer 1

from your solution manual

Answer 2

yes it is true, we can replace expression f(n) with f(x) , prove the result using Lhopital's using x, then go back to f(n).

Answer 3

that is because the continuous variable x includes integer values n

Answer 4

sometimes a solution manual will be lazy and omit this step of converting n to x , and use Lhopital's right on the expression in n. this is technically incorrect, since Lhopital is a result for functions f(x) on x reals

Answer 5

Then it is incorrect to just apply l'hopital to nth terms because you can't apply that rule to a discontinuous function, which is really what a sum of sequences is...

Answer 6

well, it is not incorrect to do it since if you convert n to x, you get a continuous function,

Answer 7

its just sloppy

Answer 8

You have to first determine which form it fits first, right? if it's in the indeterminant form of 0/0 or \(\ \frac{\infty}{\infty}\)?

Answer 9

also it doesn't matter if theres a few discontinuities, as long as f(x) is well behaved near the point x = a , or x-> oo

Answer 10

Well, technically no, but I have a nazi instructor who doesn't let us make the obvious associations like that, so I understand that it is saving a little work, but this lady is a calculator that has to see it right or her motherboard starts sparking

Answer 11

lets look at an example from your solution manual, its easier to comment on specific examples

Answer 12

then i take it the solution manual was not written by your instructor :)

Answer 13

No, that isn't necessary, I understand now. Either take the infinite limit by getting n out of numerator or convert to function and apply l'hopital for indeterminate forms

Answer 14

right you have to be careful , here is an example

Answer 15

So before you use L'Hospital's rule, you have to verify and understand the conditions of the functions of \(\ f\) and of \(\ g\)

Answer 16

lim ln(n) / n^2 , as n->00

Answer 17

on the other hand, Lhopital's rule cannot be extended to sequences which have factorials

Answer 18

That looks like it would have to be by parts...

Answer 19

we can look at f(x) = ln(x) , and g(x) = x^2

Answer 20

limit ln(x) / x^2 , as x->oo

Answer 21

that is a continuous function for x > 0

Answer 22

Does it not converge to 0?

Answer 23

if you use Lhopital you have (1/x) / (2x) = 1/ (2x^2)

Answer 24

correct

Answer 25

here is some reading , on the validity of this
http://math.feld.cvut.cz/mt/txta/2/txe3aa2f.htm

Answer 26

So passes nth term.... But we have been taught, regarding the sums of infinite series that the index can be negative anything to positive infinity as long as it eventually converges

Answer 27

My instructor she does three hours of proofs, and half of one example twice a week. I have truly learned nothing from this person. Patrick JMT has taught me all of calc II up until this point. She spends more time trying to teach us how to do proofs than solve these particular problems because if we go far enough in math we have to do proofs... I get off the train at diff eq so she is just making my life miserable

Answer 28

you would use the nth term test *if* you wanted to know the series (sum) of ln(n)/n^2 , n =1 .. oo

Answer 29

no, you can't prove the sum with that test... I thought it could only prove convergence or divergence

Answer 30

right, it demonstrates divergence only

Answer 31

if your series does not pass the nth term test, then it diverges.

if your series does pass the nth term test, then it could either converge or diverge

Answer 32

more work needs to be done

Answer 33

Yeah, however as half of the Alternating series test, the nth term test can prove convergence if both halves of the test pass... technically

Answer 34

for example the harmonic series
series 1/n n=1..oo , passes the nth term test, but still diverges

Answer 35

yeah, divergent p series

Answer 36

This might help you with some of the series tests. http://prntscr.com/54pcx4

Answer 37

sorry, what do you mean by half the alternating series test?

Answer 38

And here's one more. http://www.math.uh.edu/~leonhard/Notes/convergence_Handout.pdf

Answer 39

Well, I am sure I will be back for some taylor and Mclaurin series questions.... I just meant that the lim of a sub n has to pass the nth term test as the first rule of the alt series test, so if it passes both those requirements, then technically as far as the alt series test goes it does prove it's half of the test for convergence

Answer 40

What I did to understand series and sequences to see if they converge or diverge is just memorized the summary sheet of convergence and divergence tests... then i fit the problem to the format and calculated my answer.

Answer 41

Thanks Jhannybean, that is actually the back of my cheat sheet, which usually those things are useless on her tests.

Answer 42

Once you learn the forms though... it's easy to associate the problem to find whether the series is converging or diverging. :) Heh.

Answer 43

Well, that seems like a straightforward concept except this lady loves making the tests out of the problems at the back of the homework section that have the worst integrals, and the most insane solutions.

Answer 44

Anyhow, thanks for your help guys. I can't wait for Tuesday to see what the T1000 has cooked up for the test on this section.

Answer 45

Good luck!

Answer 46

Thanks!