# Determining convergence/divergence?

1. Apr 27, 2012

### man0005

1. The problem statement, all variables and given/known data
For (9 + cos (n))/ n^2

2. Relevant equations

3. The attempt at a solution
would this be correct?
since -1 < cos x < 1
then

8/n^2 < (9 + cos (n))/ n^2 < 10/n^2?
So use p-series from there?
Therefore it converges
Am I right?

2. Apr 27, 2012

### vela

Staff Emeritus
What exactly are you trying to determine, whether the sequence
$$a_n = \frac{9 + \cos n}{n^2}$$ converges or whether the series
$$\sum_{n=1}^\infty \frac{9 + \cos n}{n^2}$$ converges?

Last edited: Apr 27, 2012
3. Apr 27, 2012

### man0005

sorry for not clarifying
the series
where n = 1

4. Apr 27, 2012

### vela

Staff Emeritus
It looks like you're trying to use the squeeze theorem, but that's for showing a sequence converges. You want to use a test that shows that a series converges. You've correctly deduced that the terms of the series are less than or equal to 10/n2, which is a p-series that converges. Given this info, what test can you use to conclude that the original series converges?

5. Apr 27, 2012

### man0005

Would it be the comparison test?

6. Apr 27, 2012

### vela

Staff Emeritus
Yes, that'll work.

7. Apr 27, 2012

### man0005

ah thanks heaps!
just one more thing
what exactly is the meaning of the "n" value?
Would my answer be different for n =2 or n = 0?

8. Apr 27, 2012

### vela

Staff Emeritus
I assume you're referring to what appears below the summation sign. What's written there tells you the initial value of n. n=0 won't work because you'd be dividing by 0 in the first term of the series. (I fixed the error in my previous post.) You should read up on summation notation if you have further questions.

As far as convergence goes, where the series starts doesn't really matter. Whether a series converges or diverges depends solely on the behavior of the terms as n goes to infinity.

9. Apr 27, 2012

### man0005

ah i see, that clears it up
thanks again!

10. Apr 27, 2012

### man0005

Would this be a valid method of testing?
1) Ratio test to end up with
(5n^n) / (n+1)^(n+1)

2) Take out 5 / (n + 1) as a factor

3) Left with 5 / (n + 1) * 1 / (n + 1)^n

4) Since the limit of 5 / (n + 1) = 0
Using limit laws, the whole thing must therefore = 0

Since 0 < 1
It converges

11. Apr 27, 2012

### vela

Staff Emeritus
This isn't correct. You seem to have dropped the nn that was in the numerator.

12. Apr 28, 2012

### man0005

oh yeah
but wouldnt that make it

5 / (n + 1) * n^n / (n + 1)^n

So would my step 4 still be valid?

13. Apr 28, 2012

### vela

Staff Emeritus
As long as you can show that $\displaystyle\lim_{n \to \infty} \frac{n^n}{(n+1)^n}$ exists.

14. Apr 28, 2012

### man0005

hmmm not sure how to do that :S
Any hints?
Or is there an easier test that I should be doing...

Edit: Just tried it and got 1/e
Working out =
lim (n→∞) [ n / ( n + 1 ) ]ⁿ

= lim (n→∞) { 1/ [ ( n + 1 ) / n ] }ⁿ

= lim (n→∞) { 1 / [ 1 + (1/n) ] }ⁿ

= 1 / { lim (n→∞) [ 1 + (1/n) ]ⁿ }

= 1 / e (Using known identity)

That seem okay?

Last edited: Apr 28, 2012
15. Apr 28, 2012

### vela

Staff Emeritus
Yup, perfect!

16. Apr 28, 2012

### man0005

hmm just tried this question from my texbook:
Series of
3 / (3n*(ln (n))^0.5)
where n = 2

Used the integral test and found that it diverges, but answer at the back of the book says it converges. Is the answer wrong? Fairly sure I havent made any mistakes...but I can post up my working if it's meant to converge

17. Apr 28, 2012

### vela

Staff Emeritus
The book's wrong.

18. Apr 28, 2012

### man0005

Ah cool
Just to confirm the method:
1) Show that related function is positive
2) Show that it is also decreasing via negative gradient
3) Get a definite integral from 2 to infinity or 2 to N

Should end up with limit of 2(ln(N))^0.5 - 2(ln(2))^0.5 with N ---> Infinity

Second is a constant, so can remove it
Now we have lim (n→∞) { 2(ln(N))^0.5}
Which is infinity, hence it diverges

Therefore original series will diverge as well by integral test

19. Apr 29, 2012

### vela

Staff Emeritus
Yes, that's fine.