# Homework Help: Questions about convergence of series.

1. Nov 8, 2009

### Nerdynerd

1. The problem statement, all variables and given/known data
Hey, I need some help with these three problems which have to do with the convergence or divergence of the series. If you can, please let me know how you solved each problem including which kind of test you used for example. Anyways, here are the problems:
1) $$\Sigma$$ (1)/(n + n*(cosn)^2), from 1 to infinity
2) $$\Sigma$$ (5^n)/(3^n + 4^n), from 1 to infinity
3) $$\Sigma$$ (n*sin(1/n)), from 1 to infinity

Thank you in advance for your help. The soonest you can reply to me, the best because I need these answers soon! :)

2. Relevant equations
N/A.

3. The attempt at a solution

For the first one, I tried to use the integral test but I don't really think that this can be the best approach.
For the second one, I tried using the ratio test but I don't know how to factor out the terms I need to in order the result of the limit.
For the third one, I got that the series diverges because at n -> infinity, sin(1/n) ~ 1/n. Therefore, when I substitute sin(1/n) for 1/n and use the nth term test I get that the limit as n approaches infinity = 1 which is not 0, therefore, according to the nth term test the series diverges.

2. Nov 8, 2009

### lanedance

1) think about a comparison test
2) show your work & can have a look (other options here:
- does a_n go to zero? (use comparison again to simplify)
- use comparison test first to simplify, then look at ratio or root test
3) you argument sounds reasonable to me

Last edited: Nov 8, 2009
3. Nov 8, 2009

### lanedance

expanded above

4. Nov 8, 2009

### Nerdynerd

Ok, I will show you where I get stuck at the second one. And for the first one, do you mean to compare it to 1/n? I mean, (cos(n))^2 is bounded, so...
But anyways, let me post where I get stuck at #2.

5. Nov 8, 2009

### lanedance

sounds like a good idea to me
no worries, i put some extra ideas up for number 2 that may help as well

6. Nov 8, 2009

### Nerdynerd

For the second one I set: (using ratio test)

a_n = (5^n)/(3^n + 4^n)

a_n+1 = (5^(n+1))/(3^(n+1) + 4^(n+1)) which we can write as
(5*5^n)/(3*3^n + 4*4^n)

then, I take the limit of(a_n+1 / a_n) as n approaches infinity, therefore I get

lim [ (5*5^n)/(3*3^n + 4*4^n) * (3^n + 4^n)/(5^n)] instead of dividing, I multiplied by 1/a_n. 5^n cancels out and I get:

lim [ 5*(3^n + 4^n) / (3*3^n + 4*4^n)]

This is where I'm stuck. If I can find a way to cancel out a few terms and get the result of this limit I will be fine! But I don't really know how to do that.... :/

7. Nov 8, 2009

### lanedance

so before you even start have a think about using the comparison test to find a suitable series:
$$\frac{5^n}{3^n+4^n}>???$$
note its the (4^n + 3^n) thats getting in the way, how could you simplfy that for a ratio test & satisfy above?

8. Nov 8, 2009

### Nerdynerd

Hmmm... good idea.... So you mean comparing this to 5^n/4^n ? Because as we approach infinity the 4^n term becomes bigger and more significant than the 3^n. So I should compare the original one to the 5^n/4^n, right?

9. Nov 8, 2009

### Nerdynerd

Umm.... I think I meant to say 5^n/3^n instead of 5^n/4^n....

10. Nov 8, 2009

### Nerdynerd

Actually, I think I am wrong in both cases....

11. Nov 8, 2009

### lanedance

12. Nov 8, 2009

### lanedance

hint (you should end up with a factor of 2)

13. Nov 8, 2009

### Nerdynerd

Hmmm.... now I am even more confused.... If I used the comparison test and compare the initial one with 1/4^n then.... I don't know whats going on... :/

14. Nov 8, 2009

### lanedance

I think 2) diverges, so you want find a series that diverges, that is every where smaller than the one you have (hopefully i haven't missed anything)

fill in the question marks
$$\frac{5^n}{3^n+4^n}>\frac{5^n}{???+???}$$

then decide to show if a_n goes to zero, or a ratio test

15. Nov 8, 2009

### Nerdynerd

so, do you mean replacing the question marks with 3*3^n + 4*4^n ? How is this going to work though? Can I use two tests at the same time or something?

16. Nov 8, 2009

### lanedance

so noticing
3^n < 4^n

then
3^n + 4^n < 4^n +4^n

and if 0<a<b, then 1/a>1/b>0

then how to fill in the following...
$$\frac{5^n}{3^n+4^n}>\frac{5^n}{???+4^n}$$

17. Nov 9, 2009

### Nerdynerd

So, the question marks are supposed to be filled with 4^n? So we can have 2*4^n, right?

18. Nov 9, 2009

### Nerdynerd

And then, b_n is a geometric series which diverges because |r| = 5/4 = 1.25 > 1, therefore it diverges which means that a_n diverges too!!!!!!!!!!!

Let me tell you something, YOU ARE A GENIUS!!! Thanks for helping me! :)
If I can help by any means in the future let me know, even though you are way smarter than me, so I don't know if I can be any help to you! lol