# Thread: Limit, problem regards Sequence and Series

1. ## Limit, problem regards Sequence and Series

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 {an} given by an = n x IN(n/(n-1))

Determine the limit as n approach infinity, if it exists or explain why the sequence diverges.

an = 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.

2. ## Re: Limit, problem regards Sequence and Series

Hi Junks,
For your first question, try calculating the nth term of the sequence for some large n's, say n=1000, 10000, etc. You'll find that these values are not close to 0. So your analysis must be wrong. (If you don't have a calculator, use an online calculator.)

The attachment gives some hints:

3. ## Re: Limit, problem regards Sequence and Series

Originally Posted by junkwisch
an = 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
You cannot split this limit into a product of limits. This is an indeterminate form ∞ x 0. (You are right that $\lim_{n\to\infty}\ln(n/(n-1))=0$.) You have to use either Taylor expansion of $\ln(n/(n-1))$ to see how fast it approaches 0 or L'Hopital's rule. In the first case, $\ln(1+x)=x+o(x)$ when $x\to0$, so $\ln(n/(n-1))= \ln(1+1/(n-1))= 1/(n-1)+o(1/(n-1))$. Therefore, $\ln(n/(n-1))$ tends to 0 as $1/(n-1)$. Thus, $\lim n\ln(n/(n-1))= \lim n/(n-1)+o(n/(n-1)) =\lim n/(n-1)=\dots$. I'll let you come up with the final answer.

In the second case, you need to represent $n\ln(n/(n-1))$ as $\frac{\ln(n/(n-1))}{1/n}$ because L'Hopital's rule applies to ratios. Then take derivatives of the numerator and denominator and find the limit of their ratio.

Originally Posted by junkwisch
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)!)
Use the ratio test.

Originally Posted by junkwisch
I'm very sorry that I do not know how to do the nice mathematics expression.
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.

4. ## Re: Limit, problem regards Sequence and Series

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 )

5. ## Re: Limit, problem regards Sequence and Series

Originally Posted by junkwisch
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.
The correct answer to the first problem is 1. See WolframAlpha.

Originally Posted by junkwisch
lim (n!)^3/(3n)! = lim ((n+1)^3!/(3n+3)!) x (3n)!/n^3!)
Why do you claim that these limits are equal? The first is the limit of the terms an of the series. The series converges, as you'll see, so an → 0 (nth term test). The second is the limit of an+1/an. 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.

Originally Posted by junkwisch
since n^3! and (3n)! cancle each other out
Why? Not at all. Since (n + 1)! = n! * (n + 1) and (3(n + 1))! = (3n)! * (3n + 1)(3n + 2)(3n + 3), we have

$\frac{((n+1)!)^3}{(3(n+1))!} \cdot\frac{(3n)!}{(n!)^3} = \frac{(n!)^3(n+1)^3}{(3n)!(3n+1)(3n+2)(3n+3)} \cdot\frac{(3n)!}{(n!)^3}$

Cancel appropriate terms and evaluate the resulting limit.

6. ## Re: Limit, problem regards Sequence and Series

I'm so sorry emakarov , the equation for the first question is n x log(n/(n+1)) I accidentally type n-1 instead of n+1