Do Infinite Series and Sequences Converge or Diverge?

[fubag]

Hi I have a few problems with regard to infinite series and sequences:

1.) cos(n) / (n^2) summation from n = 0 to infinity

2.) cos(pi*n)/n summation from n = 0 to infinity

3.) sin(n) summation from 0 to infinity

I have to tell whether they absolutely converge, conditionally converge, or diverge.

I tried using the ration/root/limit test for them but nothing seemed to help.

any feedback will be greatly appreciated.

 

[morphism]

The "comparison test" and the "divergence test" will be useful. For the second one, in particular, knowing what cos(pi*n) looks like is also helpful.

By the way, for the first two the summation can't possibly start at n=0.

 

[mathwonk]

there are two basic limits we an compute, geometric series, and improper integrals. all other series must be compared usually to one of these. a necessary condition is that the terms go to zero, which is sufficient if their size goes to zero monotonically and they have alternating signs.

compare 1 above to the series 1/n^2 which is then compoared to the integral of 1/x^2.

2 is the other test , alternating test.

3 i am not sure of. but suspect it does not converge by the test of whether the terms approach zero.

 

[Gib Z]

My brain is exploding.

True, the series diverges by several tests, most obviously by the n-term test mathwonk mentioned. But if we go about evaluating the result;

S = \sum_{n=1}^{\infty} \sin (n) = \sum_{n=1}^{\infty} \mathRR{Im} (e^{in}) = \mathRR{Im}\sum_{n=1}^{\infty}(e^{in}) = \mathRR{Im} (\frac{1}{1-e^i}) = \frac{1+ \cos 1}{2 \sin^2 1}.

What did i do wrong :(

 

[rudinreader]

The geometric series doesn't converge when |z|=1:

s_n = \sum_{i=0}^n z^n = \frac{1}{1-z} - \frac{z^{n+1}}{1-z}, z not equal 1.

It's bounded, but is not a Cauchy sequence when |z| = 1.
(s_{n+1}-s_n = e^{i \theta n}).

But as mentioned above, the main point is that the terms z^n don't go to zero as n -> infty when |z| = 1.

\sum c_n converges \Rightarrow s_n is Cauchy \Rightarrow |c_n| = |s_n-s_{n-1}| \rightarrow 0.