Sigma (-1)^n sin(Pi/n) compare with Signma(-1)^n cos (Pi/n)
What do the terms look like as n->infinity?
With calculator, I can see the difference.
But how do you analyze this without using calculator?
As n gets large, pi/n goes to zero.
For sin -0+1-sqrt(3)/2+sqrt(2)/2 ------>althernating series ---->conver.
con -1+0-1/2+sqrt(2)/2-sqrt(3)/2 ---------increasing---diver.
oh..i see now.
So the lim for Sin(pi/n) = 0-->can't be determined...if con or div.
lim for Cos (pi/n) = 1 by div. test, it's div.
Uh? Sorry, what the heck is this all about? I've not understood a single word of what's been typed.
To check Sigma (-1)^n sin(Pi/n) and Signma(-1)^n cos (Pi/n) are divergent or convergent.
Analyzing it without calculator.
Actually you can say whether the sin series converges because it is an alternating series.
And that bears what relation to your first post? We are not psychics. Post full questions.
For the question, as has been noted, one has terms that do not tend to zero, so the sum can't converge, and one passes the alternating series test.
Note that it is impossible to work out the answer *with* a calculator.
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