Root Test for Series
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Great video! I have a test on Tuesday and your videos have been so helpful.
I'd just like to let you know, though, that this video's not on your website.
Thanks again!
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3^-n = 1/3^n. As n approaches infinity the denominator gets bigger therefore limit = zero
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Thank you for your videos! Everything is so clear.
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Erm.
@ 5:27
.... Doesn't 3^(-n) approach 1, not zero?
Otherwise it would be infinity/0...?
SIDENOTE:
Thanks for all the info. Your tank work video has helped me study for my final, the most. I've been a bit iffy on the work problems, before. /Really iffy. >>'
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i love you! thank you so much. ive been watching your videos all day for my final tomorrow. i think ill pass it because of you! :)
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I LOVE YOUR TUTORIALS!!!! THANK YOU. I'M SO GONNA ACE MY CAL TEST!!!!!! :DDDDDD
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We left-handed humans are far superior in mathematics and physics...
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just started studying this in calc II and your video was very helpful. sound quality was good and you made this topic less frightening!
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patrick you rock man
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Question:
If i have an alternating series [(-1)^n-1]/2^n
This is negative one raised to the n-1 and all of that divided by 2raised to the n power. Can i use the Root Test for that or should i used the alternating series test better? Is it incorrect if if i use the root test for this?
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God bless you dude! You have no idea how much these videos have helped me for my cal 2 final. the limit of my appreciation for your work approaches infinity. Great work!
rahlity 8 months ago 40
@christalwang it's 3^0, not 0. 3^0 equals 1, so the 1x27 = 27
robert19w 10 months ago 2