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Series question

  • Thread starter daniel_i_l
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daniel_i_l
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1. 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.
 

Answers and Replies

Dick
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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.
 
daniel_i_l
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daniel_i_l
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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?
 
Dick
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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).
 
HallsofIvy
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1. 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?
 

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