Whether the sequence Converges

[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= 0.322I'm not sure how to continue from this point forth. How to show whether or not the limit converges.

 

[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?

 

[LCKurtz]

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.

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

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

 

[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.

 

[Bohrok]

How do you know it diverges to -∞?

 

[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

 

[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.

 

[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

 

[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.

 

[Mosaness]

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

 

[tt2348]

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.

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

 

[Mosaness]

We haven't been taught that yet

 

[tt2348]

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

 

[Mosaness]

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

 

[tt2348]

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

 

[Mosaness]

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


lim_{n →∞}{ln\frac{1}{n}}

lim_{n →∞}{ln\frac{\frac{1}{n}}{\frac{n}{n}}

lim_{n →∞}{ln\frac{1}{n}}

Which can also be written as

lim_{n →∞}{ln\frac{1}{n}} = ln 1 - ln n

lim_{n →∞}{ln\frac{1}{n}} = - ln n

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

 

[Mosaness]

@ KGmphysdurham,

OP is working with a sequence, not a series !

Is my answer correct?

 

[HallsofIvy]

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

"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/∞"!)

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

So the sequence converges to 0? This seems wrong

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

 

[Curious3141]

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


lim_{n →∞}{ln\frac{1}{n}}

lim_{n →∞}{ln\frac{\frac{1}{n}}{\frac{n}{n}}

lim_{n →∞}{ln\frac{1}{n}}

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

Which can also be written as

lim_{n →∞}{ln\frac{1}{n}} = ln 1 - ln n

lim_{n →∞}{ln\frac{1}{n}} = - ln n

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

The rest is fine.