For comparing series with the direct comparison test (DCT), the definition of the test states 0 < a_n ≤ b_n, so really the equals could always be included, we just typically aren't comparing series where equality could happen for a value of n, so we just omit it and keep it simple with < or >. In the case of the problem you are referring to in the video, I suppose I included the equals since it comes into play when considering the outputs of sin(n). Specifically, the statement I write at around 15:56. Since -1≤sin(n)≤1 which means |sin(n)|≤1, I think it makes sense to include the equals in our comparison statement, since we are making conclusions from it based on the outputs of sin. Does that make sense? Let me know!
if we have same scenario of comparison test, where we have p series converges, but with comparison, it diverges. What is the series become to? conditionally convergent or we need to run more test?
You might have to give me an example of the scenario you are referring to. If i understand your question correctly, it all comes down to the comparison test you use. From the direct comparison test (DCT), if you choose to compare to a convergent p series, you need to show that the comparison series is bigger than your original series. If that does not happen, you cannot conclude anything and need to pick a different comparison series. I'd recommend reviewing the DCT, that may answer your question. It is lesson 25 in my calc 2 playlist. Hope this helps!
@@JKMath ok thanks, i have another question, lets say, i have a problem where i used integral test to find |an| which i got convergent, do i need to do another test to check if its absolutely convergent?
@@Rexcel-y6h No, referencing the chart at 11:22 in the video, if you can show the the series for |an| is convergent, no matter the test, then you can conclude that the series for just an is convergent as well. You can conclude that the series is absolutely convergent.
thank you, i got a 94 on my integral exam :). keep doing what you're doing!
Awesome! Nice job on the exam :) Will do!
Thank you! This video was really helpful!
Why did you include the equal sign when comparing An and Bn. I thought it was either < or >. 15:09
For comparing series with the direct comparison test (DCT), the definition of the test states 0 < a_n ≤ b_n, so really the equals could always be included, we just typically aren't comparing series where equality could happen for a value of n, so we just omit it and keep it simple with < or >. In the case of the problem you are referring to in the video, I suppose I included the equals since it comes into play when considering the outputs of sin(n). Specifically, the statement I write at around 15:56. Since -1≤sin(n)≤1 which means |sin(n)|≤1, I think it makes sense to include the equals in our comparison statement, since we are making conclusions from it based on the outputs of sin. Does that make sense? Let me know!
@@JKMath Makes sense, thanks
Thanks
Please keep it up
if we have same scenario of comparison test, where we have p series converges, but with comparison, it diverges. What is the series become to? conditionally convergent or we need to run more test?
You might have to give me an example of the scenario you are referring to. If i understand your question correctly, it all comes down to the comparison test you use. From the direct comparison test (DCT), if you choose to compare to a convergent p series, you need to show that the comparison series is bigger than your original series. If that does not happen, you cannot conclude anything and need to pick a different comparison series. I'd recommend reviewing the DCT, that may answer your question. It is lesson 25 in my calc 2 playlist. Hope this helps!
@@JKMath ok thanks, i have another question, lets say, i have a problem where i used integral test to find |an| which i got convergent, do i need to do another test to check if its absolutely convergent?
@@Rexcel-y6h No, referencing the chart at 11:22 in the video, if you can show the the series for |an| is convergent, no matter the test, then you can conclude that the series for just an is convergent as well. You can conclude that the series is absolutely convergent.
@@JKMath ok thank you