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Topic: sin n over n
Replies: 10   Last Post: Aug 6, 2011 2:47 PM

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Rob Johnson

Posts: 1,771
Registered: 12/6/04
Re: sin n over n
Posted: Jun 28, 2011 11:20 AM
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In article <2b665f67-a945-419f-ab36-812899f6c4b0@bl1g2000vbb.googlegroups.com>,
Babu <tprof@hotmail.it> wrote:
>It is not my homework :-)
>I am not a student, but a retired Italian high school teacher
>I love trying to solve puzzles and math problems
>Usually they have a solution in the end of the book.
>
>But this one was invented by me and I am not sure if the solution
>works
>I put it here
>http://img855.imageshack.us/img855/5307/sinnovern.png
>
>Any help, suggestion etc will be highly appreciated


While your result is true, the series does converge conditionally,
your proof needs modification.

It is true that |a_k| <= b_k and that b_k > b_{k+1}; however, that
does not imply that a_k > a_{k+1}, nor even that |a_k| > |a_{k+1}|.

You sum together 3 consecutive terms of b_k, and call it a_k. Not
only does this confound what you mean by a_k, but those elements
grouped together don't always have the same sign, nor do they cover
all of the b_k (e.g. b_25 is not covered by a_k; a_7 covers b_22
through b_24 and a_8 covers b_26 through b_28).

The proper way to handle this is to show that

N
---
> sin(n)
---
n=1

is bounded independent of N. Because 1/n is monotonic decreasing,
the series converges conditionally by Dirichlet's convergence test
<http://en.wikipedia.org/wiki/Dirichlet%27s_test>.

Rob Johnson <rob@trash.whim.org>
take out the trash before replying
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