Proving Divergence of |cos(n)a_n| w/ Converging Series

In summary, if the series (cos(n) a_n) converges then so does the series (cos(n) a_n)^2 - 1. However, if the series (cos(n) a_n)^2 - 1 converges then cos(n)^2 doesn't have bounded partial sums, proving the series (cos(n) a_n) diverges.
  • #1
daniel_i_l
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Homework Statement


Lets say that I have some sequence [tex](a_n)[/tex] which converges to 0 at infinity and that for all n [tex]a_{n+1} < a_n [/tex] but the sequence [tex](a_n)[/tex] diverges. Now I know that the series
[tex](cos(n) a_n) [/tex] converges but can I use the following argument to prove that
[tex]|cos(n) a_n| [/tex] doesn't converge:
[tex] |cos(n) a_n| >= {cos}^{2}(n) a_n = {a_n}/2 + {(cos(2n)) a_n}/2 [/tex]
And since [tex]{(cos(2n)) a_n}/2 [/tex] converges and [tex] {a_n}/2 [/tex] diverges
[tex] {cos}^{2}(n) a_n [/tex] diverges and so [tex] |cos(n) a_n| [/tex] diverges.
Is that always true?
Thanks.
 
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  • #2
That seems ok, provided you can prove cos(2n)*a_n converges. Are you using Abel-Dedekind-Dirichlet (summation by parts)? I just figured out how it works, so I had to ask.
 
  • #4
  • #5
Well I know that cos(n) has bounded partial sums and cos(2n) = 2(cos(n))^2 - 1 so that means that cos(2n) also has bounded partial sums right?
 
  • #6
daniel_i_l said:
Well I know that cos(n) has bounded partial sums and cos(2n) = 2(cos(n))^2 - 1 so that means that cos(2n) also has bounded partial sums right?

Not at all! cos(n)^2 doesn't have bounded partial sums and neither does 1! How did you show cos(n) had bounded partial sums?? Try and apply the same technique to cos(2n).
 
  • #7
daniel_i_l said:

Homework Statement


Lets say that I have some sequence [tex](a_n)[/tex] which converges to 0 at infinity and that for all n [tex]a_{n+1} < a_n [/tex] but the sequence [tex](a_n)[/tex] diverges.
You mean series rather than that last "sequence", right?
 

What is the purpose of proving divergence of |cos(n)a_n| with converging series?

The purpose of this proof is to show that the series |cos(n)a_n|, where a_n is a converging series, also diverges. This is important in understanding the behavior and properties of trigonometric series.

What is the definition of divergence in a series?

Divergence in a series means that the sum of the terms in the series does not approach a finite limit as the number of terms increases. In other words, the series does not have a definite value and instead grows infinitely.

How do you prove divergence of a series?

To prove divergence of a series, you must show that the terms in the series do not approach zero or that the series does not satisfy the conditions for convergence, such as the divergence test or comparison test. In this case, we will use the comparison test to show that |cos(n)a_n| diverges.

What is the comparison test in series convergence?

The comparison test is a method used to determine the convergence or divergence of a series by comparing it to a known series. If the known series converges and the terms in the original series are smaller than or equal to the terms in the known series, then the original series must also converge. If the known series diverges and the terms in the original series are larger than or equal to the terms in the known series, then the original series must also diverge.

How does the comparison test apply to proving divergence of |cos(n)a_n| with a converging series?

In this case, we will use the known series 1/n^2 as our comparison series. We can show that |cos(n)a_n| is always greater than or equal to 1/n^2, which means that if 1/n^2 converges, then |cos(n)a_n| must also diverge. Since 1/n^2 is a known converging series, this proves the divergence of |cos(n)a_n|.

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