Limit Comparison Test and Direct Comparison Test
Uploader Comments (patrickJMT)
Top Comments
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@asianconfusion many years of making videos through many moods and levels of health.
All Comments (144)
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@patrickJMT hahaha thats so evil. Thanks for making this easier buddy
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You are so much more clear than my teacher, thank you!
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Nice.
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Was there ever a video on p-series?
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it's about time he got some commercials
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I am very happy to see the vidoe Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges from you, hopefully the others also are happy for You
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I Love The Video It Can Increase My Knowledge Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges
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Steady I Really Like This Video Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges
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Nice Video That You Share , So Very Nice Thanks You Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges.
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I Really Like The Video From Your Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges
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Thank you Patrick.
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@patrickJMT At the beginning, you said:
[1 + sin(n)] / [10^n] is less than or equal to 2/10^n, "for all n is greater than or equal to 1". Why did you choose 1? Isn't it 0, because that's what our original series had n equal to?
Thank you for your help! Your videos are great!
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For those wondering about the 2/(n^2) expression.
Consider that the expanded terms he showed are constantly multiplying by something *smaller than one*; thus, the sum would constantly be getting smaller. If you remove these terms making your sum smaller, and just keep the final two terms, you will then have a series that is in fact larger than the initial series given.
He also keeps the final two because together they make up a p-series, which is easy to tell divergence/convergence.
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Once again I get stuck on a problem and magically ur video is on the same one! Thanks sir
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Nice
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i think the ratio test works better for the last one.
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Thanks a lot for the very 1st problem of this video. I had the same one in my Hw.:D
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Omg! I did not understand why only (2/n) and (1/n) were used until now! My math teacher showed this example a month ago and I remember something being negative ^^". Either my memory sucks or that explanation failed, but I wasn't the only one extremely confused that day.
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Nevermind I figured it out. :)
I don't get it...
khilozozo02 1 week ago
@khilozozo02 what dont you get
patrickJMT 6 days ago
how do we know when to use the direct comparison test or the limit comparison test? im confused
eternity23211 3 months ago
@eternity23211 part of the fun is figuring that out
patrickJMT 3 months ago
OKAY PATRICK I DONT GET IT!!
how did you cancel all the terms but leave the last two, the 2/n and the 1/n you said you cancelled the rest because they were smaller than one.. but so is 2/n and 1/n. Why did you leave those terms intact?
please answer quick test in 2 days :)
and thanks for all your help
floyd617 11 months ago
@floyd617 i leave those terms intact because it provides a useful expression: 2/n^2 and the series associated with that converges. you have to realize that you are trying to show the given expression is smaller than some convergent series (or larger than some divergent series). it is not mechanical to do this: how can one justify it? for this example, i found a way!
(i hope i am remember correctly what is in the video)
patrickJMT 11 months ago 2