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Re: sin n over n
Posted:
Jun 28, 2011 11:20 AM


In article <2b665f67a945419fab36812899f6c4b0@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 to view any ASCII art, display article in a monospaced font



