Hi, its me again
I got some problems with limit of the series and sequence and here are the question I'm not so sure about and got stuck with.
The first question is
Consider for n morethanorequalto 1, the sequence {a_{n}} given by a_{n} = n x IN(n/(n-1))
Determine the limit as n approach infinity, if it exists or explain why the sequence diverges.
a_{n} = n x IN(n/(n-1))
= lim(n approach infinity) n x lim(n approach infinity) IN(n/(n-1) (IN is for log based e)
= lim(n approach infinity) n x lim(n approach infinity) IN (1/(1-(1/n))
since n approach infinity 1/n will become very small
= lim(n approach infinity) n x lim(n approach infinity) IN(1/1)
= lim(n approach infinity) n x 0
= 0
The answer I have is that the limit of this sequence is equal to 0. Is this correct?
The second question is
Using an appropriate test, determine whether the following series is convergent or divergent
ZIGMA (with infinity on top of it) and n=1 under it (ZIGMA as in series). Zigma ((n!)^3 /(3n)!)
I do not know how to do the second question. So if any one can explain to me I'll be eternally grateful until the day I died
Best Regards
Junks
I'm very sorry that I do not know how to do the nice mathematics expression.
You cannot split this limit into a product of limits. This is an indeterminate form ∞ x 0. (You are right that .) You have to use either Taylor expansion of to see how fast it approaches 0 or L'Hopital's rule. In the first case, when , so . Therefore, tends to 0 as . Thus, . I'll let you come up with the final answer.
In the second case, you need to represent as because L'Hopital's rule applies to ratios. Then take derivatives of the numerator and denominator and find the limit of their ratio.
Use the ratio test.
Not knowing how to write LaTeX is one thing, bt if you write lN for ln and ZIGMA for Sigma next time, you vIlL geT AN anSwER LiKe DIS.
thank you for your reply, i followed your step and got an answer of -1 for the first question which is the same as the answer my calculator gave me .
However, I also have a go with question 2, I manage to get an answer but since I'm not familiar with ratio test. Is it possible that you guy can check it out for me and tell me what I did wrong.
Sigma with infinity on the top and n=1 under it, Sigma (n!)^3 / (3n)!
lim (n!)^3/(3n)! = lim ((n+1)^3!/(3n+3)!) x (3n)!/n^3!)
since n^3! and (3n)! cancle each other out
lim 1!/3! = 1/6
Since L=1/6, it is less than 1 thereby the series converge
Is what I did above correct? If not please tell me what I did wrong
thankyou,
(P.S emakarov, I actually thought Sigma is spelled with a Z )
The correct answer to the first problem is 1. See WolframAlpha.
Why do you claim that these limits are equal? The first is the limit of the terms a_{n} of the series. The series converges, as you'll see, so a_{n} → 0 (nth term test). The second is the limit of a_{n+1}/a_{n}. It is considered in the ratio test. Even for converging series this limit does not have to equal zero. Besides, why did you swap factorial and cube? It should say (n!)^3 and ((n+1)!)^3, not (n^3)! and ((n+1)^3)!; these are completely different things.
Why? Not at all. Since (n + 1)! = n! * (n + 1) and (3(n + 1))! = (3n)! * (3n + 1)(3n + 2)(3n + 3), we have
Cancel appropriate terms and evaluate the resulting limit.