Absolute convergence test?

Question

Absolute convergence test


Test the series below for convergence using the Ratio Test.
(Σn=1 to ∞) (n+2)/(4^(3n+4))

The limit of the ratio test simplifies to limn→∞|f(n)| where f(n)=?
the limit is:?

Answer 1

Your question: "Absolute convergence test?

Absolute convergence test

Test the series below for convergence using the Ratio Test.

(Σn=1 to ∞) (n+2)/(4^(3n+4))

The limit of the ratio test simplifies to limn→∞|f(n)| where f(n)=?

the limit is:?"



If you know that much, it would be unreasonable that you are seeking help here. If you are copying from a text without any understanding, you are asking at the wrong place. Try here instead:



• http://www.freemathhelp.com/

• http://www.purplemath.com/



Due to the large number of irrational and socipathic posts on this site, I have concluded that it is a dead end for math help. This is not the sort of site where people exercise their intellect or attempt to explain complex mathematics. Little to no rational inquiry and analysis occurs on this site, and that statement covers all categories.



I suppose that is probably true for most so-called social media sites.



There are some persons who will supply what they believe to be the correct answer, and some of them will even show the steps they took to solve it. But if you do not understand what they are doing, or have no way of knowing if their answer is correct, it is a waste of your time to inquire.



Given the variation in the sort of math problems you ask, it is difficult for me to believe you are seriously seeking help with concepts. Most people in that position have one math class for which they are seeking help. You have not asked for enough detail in your question to lead me to believe that you are seeking more insight into the topic. I decline to give an example of what asking for enough detail means. In your case it would mean, "And how do you know if the answer is correct? And what does it mean?"



Seriously, are you writing a book? Or does Yahoo! pay you to post this stuff?



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Answer 2

f(n)=(n+3)/[4^3(n+2)

lim n-->∞ f(n)=1/64 which <1 so series

converges.

Answer 3

lim n→∞ ((n+3)/(4^(3n+7)) )/(n+2)/(4^(3n+4))

=lim n→∞ (n+3)/((n+2)4^3)

=lim n→∞ (1+3/n)/(64(1+2/n))

=1/64