need help with this, use the geometric series test, from one to inifinty of (-3/2)^n, i got it was convergent, but just want to check my work
you do know that \[\frac{-3}{2} \cancel{\in} (-1,1)\]
?
see that, is where i dont understand, how do you know that, i have looked at samples of the geo test, but everytime i use it, i get it wrong, but i use the test correctly
how do i know the number -3/2 is not between -1 and 1?
yes, i see that
oh then what does "how do you know that" refer to?
how to show with the geometric test, like how do i show my work with that, not trying to make you do the work, just how do i prove that
r has to between -1 and 1 for it to converge
r being the common ratio
otherwise it diverges
that is the test
is it really that simple?
yep
then why do teachers make it so gosh dang more complicated
example \[\sum_{n=1}^{\infty } (\frac{1}{2})^n \text{ this one converges because } \frac{1}{2} \in (-1,1)\]
example \[\sum_{n=1}^{\infty} (\frac{-1}{2})^n \text{ also converges because } \frac{-1}{2} \in (-1,1)\]
so since that it is not between -1 and 1, it diverges
http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx don't know if you ever heard of pauls notes but he has really great notes :)
right
thank you, that makes a lot more sense, looking at notes and books, this is the easiest explanation i have heard
and no, i havent
usually books usually the inequality notation maybe that seems harder to read that what the inequality actually means... when it says: \[\sum_{n=1}^{\infty}a (r)^n \text{ converges when } |r|<1 \\ \text{ the inequality is saying } -1<r<1 \\ \text{ or \in words everything between -1 and 1, exclusive }\]
exclusive meaning we are not including -1 and 1
anyways I'm glad it makes more sense now
thank you, so and pauls notes will have the other test im assuming
i'm pretty sure and i could be mistaken paul's notes has everything you will need for calculus
again thank you
np
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