Good catch. It's true that for the limit comparisons test (LCT) that the limit of a_n/b_n should be finite and positive, so it seems I made a slight error in my strategy there. The LCT is probably not the most appropriate test to use in that case, at least not the way I did it in the video. However, the answer is still correct. The series still converges. Here's some options of what I should have done instead: What I could have done first is checked for absolute convergence which says that ∑a_n converges if ∑|a_n| converges, which you could test for ∑-5/(n^2+n). You could check the convergence of ∑|-5/(n^2+n)| which would become ∑5/(n^2+n) and then that could be shown to converge using the LCT since then the limit of a_n/b_n would be finite and positive (5) assuming you choose ∑1/n² as the comparison series. Then, since ∑|a_n| converges, ∑a_n converges absolutely. So the answer is still the same, the series converges. Another way it could be done is by realizing that multiplying a series by a nonzero constant (such as -5) will not affect the convergence or divergence of the series. So if you rewrite ∑-5/(n^2+n) as -5·∑1/(n^2+n) then you could use the LCT on ∑1/(n^2+n) to show it converges. Multiplying it by -5 does not affect its convergence. My apologies for any confusion caused by my unfortunate error of strategy in the video. I hope my explanations here in this comment clear things up for you! Let me know if you have any further questions. I am always happy to help :)
Hey for 23:14 can you explain how the alternating series converges, because since the numerator is (-3), it would make the values from the left of the inequality always bigger than the values on the right
Hi! My apologies for the confusion there, it seems I made a slight error. You are right, the inequality is not true and that is my mistake. My use of the alternating series test (AST) for that series would have been correct given that a_n for the series was positive. In other words, if we were working with 3/(n+1) instead. However, we can fix this pretty easily (and this is what I should have done/showed in the video): An important fact to know about series and convergence/divergence is that multiplying them by a constant, either positive or negative, does not effect the convergence of the series. So if you pull out a negative 1 and rewrite ∑(-1)^n*(-3)/(n+1) as (-1)∑(-1)^n*3/(n+1) then you could use the AST to show it converges, as a_n for the series is now positive and the inequality for checking if the terms are decreasing will be true. Multiplying the original series by -1 ultimately will not affect the convergence of the series (you can just pull the constants outside the series!). In fact, you could even pull out the entire -3 and still get the same result. Your a_n in that case would be 1/(n+1), and that would meet the requirements of AST as well. You can always use this trick when working with a series where a_n is negative. I hope this helps clear things up, and once again my apologies on the error in explanation in the video.
Why we can use limit comparsion test in 52:38 , as -5/(n^2+n) < 0
Good catch. It's true that for the limit comparisons test (LCT) that the limit of a_n/b_n should be finite and positive, so it seems I made a slight error in my strategy there. The LCT is probably not the most appropriate test to use in that case, at least not the way I did it in the video. However, the answer is still correct. The series still converges. Here's some options of what I should have done instead:
What I could have done first is checked for absolute convergence which says that ∑a_n converges if ∑|a_n| converges, which you could test for ∑-5/(n^2+n). You could check the convergence of ∑|-5/(n^2+n)| which would become ∑5/(n^2+n) and then that could be shown to converge using the LCT since then the limit of a_n/b_n would be finite and positive (5) assuming you choose ∑1/n² as the comparison series. Then, since ∑|a_n| converges, ∑a_n converges absolutely. So the answer is still the same, the series converges.
Another way it could be done is by realizing that multiplying a series by a nonzero constant (such as -5) will not affect the convergence or divergence of the series. So if you rewrite ∑-5/(n^2+n) as -5·∑1/(n^2+n) then you could use the LCT on ∑1/(n^2+n) to show it converges. Multiplying it by -5 does not affect its convergence.
My apologies for any confusion caused by my unfortunate error of strategy in the video. I hope my explanations here in this comment clear things up for you! Let me know if you have any further questions. I am always happy to help :)
@@JKMath Thanks!
Hey for 23:14 can you explain how the alternating series converges, because since the numerator is (-3), it would make the values from the left of the inequality always bigger than the values on the right
Hi! My apologies for the confusion there, it seems I made a slight error. You are right, the inequality is not true and that is my mistake. My use of the alternating series test (AST) for that series would have been correct given that a_n for the series was positive. In other words, if we were working with 3/(n+1) instead. However, we can fix this pretty easily (and this is what I should have done/showed in the video): An important fact to know about series and convergence/divergence is that multiplying them by a constant, either positive or negative, does not effect the convergence of the series. So if you pull out a negative 1 and rewrite ∑(-1)^n*(-3)/(n+1) as (-1)∑(-1)^n*3/(n+1) then you could use the AST to show it converges, as a_n for the series is now positive and the inequality for checking if the terms are decreasing will be true. Multiplying the original series by -1 ultimately will not affect the convergence of the series (you can just pull the constants outside the series!). In fact, you could even pull out the entire -3 and still get the same result. Your a_n in that case would be 1/(n+1), and that would meet the requirements of AST as well. You can always use this trick when working with a series where a_n is negative. I hope this helps clear things up, and once again my apologies on the error in explanation in the video.