Convergence of Series with Cosine Terms

[de1irious]

How would I show that the series whose terms are given by
(cos n)/(1+n) does not converge absolutely? Thanks so much!

 

[courtrigrad]

$$\sum_{n=0}^{\infty} \frac{\cos n}{1+n}$$

So $$\sum_{n=0}^{\infty} \frac{\cos n}{1+n} \sim \frac{1}{n}$$

 

[de1irious]

Hi sorry, I'm having trouble understanding that. How am I supposed to compare that?

 

[d_leet]

How would I show that the series whose terms are given by
(cos n)/(1+n) does not converge absolutely? Thanks so much!

Is it cos(n) or cos(n*pi)? If it is the first then the limit comparison test should wor fairly well with the series 1/n.

 

[de1irious]

You mean this limit comparison test? http://mathworld.wolfram.com/LimitComparisonTest.html

But what limit does it tend to? I thought |cos n| didn't tend to a limit as n--> infinity.

 

[d_leet]

You mean this limit comparison test? http://mathworld.wolfram.com/LimitComparisonTest.html

But what limit does it tend to? I thought |cos n| didn't tend to a limit as n--> infinity.

Yea that test, cos(n) doesn't but it is bounded so I think if you use that fact and maybe the squeeze theorem you should be able to show that the series doesn't converge absolutely. It shouldn't be very hard to show that the series does converge as is using the alternating series test, but I'm not sure if it is cos(n) as opposed to cos(pi*n).