View Full Version : converge absolutely or conditionally, or diverges?
rcmango
Jan30-07, 11:23 PM
1. The problem statement, all variables and given/known data
Determine if the series converges absolutely, converges conditionally, or diverges.
equation is here: http://img409.imageshack.us/img409/7353/untitledly5.jpg
2. Relevant equations
maybe alternating series, or harmonic series?
3. The attempt at a solution
not real familiar with tan with series.
haven't tried much, need supporting work for the answer.
need help.
chanvincent
Jan31-07, 01:32 AM
As n-> infinity, tan(1/n) -> tan(0) -> 0
Does this help?
quasar987
Jan31-07, 02:01 AM
Actually, this says nothing at all about the series. The implication is one way only: "Sum a_n converges ==> a_n-->0" but "a_n-->0 ==> nothing".
Actually the series satisfies all the criteria corresponding to the convergence of an alternating series. Remains to see if it converges absolutely. I.e. does
\sum_{n=1}^{\infty}\tan(n^{-1})<\infty
??
rcmango
Jan31-07, 03:49 AM
no it doesn't converge absolutely because it continues on to infinity.
however, i do ask, how do you know to test it to be less than infinity? in other words, the convergence for a alternating series passes. but what other series convergence did not pass?
so ultimately, this will converge conditionally.
for my work, i could prove this by showing the alternating series? and then showing that it also continues on to infinity?
thanks again for all the help so far.
quasar987
Jan31-07, 04:37 AM
What do you mean by "continues on to infinity" ?
rcmango, i think you mean using the Leibniz test (for alternating series)
there are three conditions, check all to prove.
HallsofIvy
Jan31-07, 07:10 AM
As n-> infinity, tan(1/n) -> tan(0) -> 0
Does this help?
It is more to the point that tan(1/n) is decreasing. Knowing that it goes to 0 is neither necessary nor helpful.
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