Whether the sequence Converges

1. Jun 26, 2012

Mosaness

1. Write out the first five terms of the sequence, determine whether the sequence converges, and if so, find its limit.

{(In n)/n}+∞n = 1

α1 = 0
α2= 0.347
α3= 0.366
α4= 0.347
α5= 0.322

I'm not sure how to continue from this point forth. How to show whether or not the limit converges.

2. Jun 26, 2012

klondike

I assume you meant {ln(n)/n}. The numerator and denominator both grow as n grows. Which one grows faster? Can you see the boundary of the sequence as n grows?

3. Jun 26, 2012

LCKurtz

You mean whether or not the limit exists. Or, equivalently, whether the sequence converges.

Hint: How would you do it if it was $\frac {\ln x}x\ ?$

4. Jun 27, 2012

Mosaness

I am honestly not sure. I know that it doesn't converge. It diverges to -∞. I just don't know how to show it. Natural logs or logs have never been my strong suit.

5. Jun 27, 2012

Bohrok

How do you know it diverges to -∞?

6. Jun 27, 2012

Mosaness

I was doing some reading up on this and it said to use L'Hospital's Rule. Which pretty much states that lim (n -> ∞ )$\frac{f(x)}{g(x)}$ = lim (n -> ∞)$\frac{f'(x)}{g'(x)}$.

So, you divide both the numerator and the denominator by x (because x is the highest power in the denominator), which gives you ln(1/x)/x

This is the same as ln 1 /x
= ln 1 / ∞
= 0/∞
= 0?

So the sequence converges to 0? This seems wrong

7. Jun 27, 2012

DrSuage

Look up convergence tests in a book.

Integral test, Ratio test, Comparison test.
once you understand what the idea of these things are you will be able to do this question.

8. Jun 27, 2012

tt2348

Think of it as ln(n)/n=ln((n)^(1/n))... What is the limit as n^(1/n) goes to infinity? The plug that into ln

9. Jun 27, 2012

Mosaness

I do apologize. I was looking at a similar question, {ln$\frac{1}{n}$}.

Working out this problem showed that:

term #1 (n = 1) is 0
term #2 (n = 2) is -0.693
term #3 (n = 3) is -1.099
term #4 (n = 4) is -1.386
term #5(n = 5) is -1.609

Taking the limit o this shows:

lim ln$\frac{1}{n}$
= lim ln$\frac{1/n}{n/n}$
= lim ln ln$\frac{1/n}{1}$
I'm not sure how to proceed further.

10. Jun 27, 2012

Mosaness

As for the previous question, the limit converges to 0.

11. Jun 27, 2012

tt2348

Those are for series convergence criteria.. Sans the comparison test

12. Jun 27, 2012

Mosaness

We haven't been taught that yet

13. Jun 27, 2012

tt2348

Oh! Ln(1/n)=ln(1)-ln(n)=-ln(n)...

14. Jun 27, 2012

Mosaness

Diverges. To - infinity. Correct? And for my original question, that converges to 0. I am sorry

15. Jun 27, 2012

tt2348

Since ln(n)->inf, -ln(n)->-inf

16. Jun 27, 2012

Mosaness

Here is what I have gleamed, and understood. And I appreciate the patience you all showed while I tried to comprehend it.

limn →∞{ln$\frac{1}{n}$}

limn →∞{ln$\frac{\frac{1}{n}}{\frac{n}{n}}$

limn →∞{ln$\frac{1}{n}$}

Which can also be written as

limn →∞{ln$\frac{1}{n}$} = ln 1 - ln n

limn →∞{ln$\frac{1}{n}$} = - ln n

limn →∞ -ln n = -∞ as the bigger that n gets, the more negative -lnn will get.

17. Jun 27, 2012

Mosaness

Is my answer correct?

18. Jun 27, 2012

HallsofIvy

Staff Emeritus
"So"??? You just said to use L'Hopital's rule which has nothing to do with "divide both numerator and denominator by x". The numerator is ln(x) and its derivative is 1/x. Then denominator is x and its derivative is 1. f'/g'= (1/x)/1= 1/x. That "goes to 0" (Do NOT write "1/∞"!)

Surely you do not believe that dividing "ln(x)" by x gives ln(1)???

19. Jun 27, 2012

Curious3141

These first two steps are completely redundant, since you just ended up where you started.

The rest is fine.