Telescoping Series
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- This calculus 2 video tutorial provides a basic introduction into the telescoping series. It explains how to determine the divergence or convergence of the telescoping series. It also explains how to use the telescoping series to find the sum of the infinite series by taking the limit as n goes to infinity of the partial sum formula. This tutorial contains examples and practice problems with factoring and partial fraction decomposition with the telescoping series.
Improper Integrals:
• Improper Integrals - C...
Converging & Diverging Sequences:
• Converging and Divergi...
Monotonic & Bounded Sequences:
• Monotonic Sequences an...
Absolute Value Theorem - Sequences:
• Absolute Value Theorem...
Squeeze Theorem - Sequences:
• Squeeze Theorem For Se...
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Geometric Series & Sequences:
• Geometric Series and G...
Introduction to Series - Convergence:
• Convergence and Diverg...
Divergence Test For Series:
• Divergence Test For Se...
Harmonic Series:
• Harmonic Series
Telescoping Series:
• Telescoping Series
__________________________________
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
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I've said this before and I'll say it again. You are a lifesaver. What you explained in 20 minutes, my professor could not in 2 hours.
Thank god Im born in a age where knowledge is so easily accessible. And thank you for teaching us. I have immense respect for people like you who spread knowledge so the rest of us become more aware.
if u where not born in this age exams would also have been easier and see competition
@@mousumimishra4741 School has definitely gotten more accessible for people as time has gone on so I don't agree with this statement.
@@4seth yeah that's what am saying since schools weren't as accesible in the past years exams were easier since less people wrote them so less competition aswell
@@mousumimishra4741 Less people wrote them because they weren't smart enough to get in without the internet. There are people who have the ability to understand calculus by reading a math textbook. The bar was much higher in previous era's.
@@4seth yeah but population has risen more than college seats have so someone who was average like 20 yrs back could get a good college relatively easily than today
Professor Organic Chemistry Tutor, thank you for an awesome explanation of the Telescoping Series Test in Calculus Two. Pattern recognition and problem solving increase my full understanding of the Telescoping Series Test. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
You deserve an award . Seriously, you are the best in all your videos
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Removing the red point left in 4:10 is so SATISFYIIING!
You deserve to be a professor actually teaching in college. My calc 2 professor barely helps me until i understand more and more
at 2:28 why did you use n-2 not n-1 or something else?
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thanks dude, you just helped me pass my calculus based astrophysics exam for the first year in Australian National University
are you literally one minute ago?
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@@cameroneinstein546 nothing
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So helpful sir. I can say after this video, I get it.
Thank you sooo much all your videos help a lot!!!
Thank you so much! My classes are online now because of COVID-19 and it's hard to understand the class material. These videos are extremely helpful.
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Can someone please tell how do we get to know when to use an -1 ,an-2 or an-3
At 19:23, is there a reason why you begin at an - 5? How far should we start to see how the terms with a variable in it cancel out?
He isn’t being remotely rigorous so there’s no hard and fast rule as to where you should start when you’re looking for a pattern like this. If you want to save time and have a result you can trust, just keep it in summation notation, break up the sum, and manipulate the index so that it cancels. Far less ambiguous that way.
No he is just making sure that he knows exactly which terms cancel and which don't you can start at any number like I did at an-3
You are a life savour ♡! Thank you
That last one seems very involved. I know this is a telescoping video, but i was trying to see if there was another way to figure out whether it converges or not, and ran into some trouble. Tried integral test but i end up with [ln(Infty)-ln(Infty)]-[ln(1/2)].
You're a savior.
This is most likely the easiest method for determining what the series converges to. You just have to be good at partial fractions
Great video MAN
Thank you for this video. It helps a lot!
are telescoping and difference series same thing in sequence and series
Thank you sooooo much you are amazing!
where does a(sub)n-5 come from? Confused
He picked a term far enough from the end of the series [a(sub)n] just to show us what is going on towards the last few terms as the series approaches a(sub)n. You can pick any a(sub)n-k (where k is any natural number) and continue the calculations as so a(sub)n-(k-1) until you reach a(sub)n-1 then ultimately a(sub)n which should be the last term.
Simple steps, nice...
Even if I had access to the biggest library, I'd probably never find out about this. I'm really glad search engines of today is so advanced my pathetic description still pointed here.
Happy to say that I understand this theorm, but some of the other theorms from your other videos... Confusing a bit
13:44 Wait why does (n+1) goes first then (n+3) shouldn't they be switched?
thank you u r the best
Thank u very much sir
i dont get it, in the previous video when sum was not equal to 0, we said it diverges. why this time when it equals to a number we said it converges??
It converges if the sum is a finite number (like 0,5,-7,...) it diverges if it is +♾️ or -infinite
thank you
Good job
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thanks
2:27 why did he take n-2
Thanku❤️
Thanks sir
I don't get the last part of question 1, why did you put a(n-2)???
He just use a(n-2) so that you could check what numbers would be cancelled in nth term term when n is infinite.
Yeah I got it thanks@@nellvincervantes3223
Welcome miss
مرحبا
تكدرين هم تحليهة بغير طريقة عادي مهم توجدين قيم a,b
Good man
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5:24 bookmark
At 4:54vwhy is the limit being subtracted? you lost me there
He's not subtracting it. The limit, infinity, is substituting "n".
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nerd
thanks btw
you deserve a noble prize for saving the lives of many students including me. Thank you a lot !
Noble Prize for what? Ok, peace.
That idea,(notion), is a bit exaggerated.
@@normanhenderson7300 No its not exaggerated he is literally saving lives out here!
Lik bleshal man
*Nobel Prize
series are literally going to be the death of me. the rest of calc 2 has been so easy up until now
for the last problem, how did you know to start with an-5?
it does not matter really. you can start with an-2, that's what i do each time. ultimately, all of those terms will get canceled other than the last term and potentially second to last term
@@erikb811 I've been looking for an answer to that the past eons thank you
He started with it just to show as an example
@@erikb811 Bro, your a life saver. I've literally been stuck on this for the whole week!
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I know you get this a lot, but good job. You really are great at explaining things concisely. Which helps those of us just learning the information for the first time. Cheers.
you are a legend.... i just dont have a knack for sequences and series but you just made my day and semester..... thank you all the way from South Africa
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My logic on the last problem is that once it starts repeating I counted at how many terms it repeated. This number was 3 [ ]'s. So then I always do 1- that many terms, and this is how I determine the exact amount of an-x terms that I would need. I also observed that it will always involve the first numbers in the bracket for the first listing and the last numbers in the brackets for the an-x listing. I know my explanation sounds confusing but hopefully, someone finds it helpful.
How do you know what number to do for An-#? Don't get why we did An-1 on one problem but An-2 on another
This guys videos are very helpful but he does this often. Very small steps in the problem he won’t explain and you see him apply it differently for different problems, but he still won’t mention it
at 2:20 could someone explain where a sub n-2 comes from? been trying to compare this video with lecture notes and cant seem to find an answer anywhere
Thank you for this. We went over telescoping series for practically 5 minutes in class out of 2 weeks and its probably going to be on the exam, yer a lifesaver.
And which class are you in?
At 2:28 can you explain how you got a(sub)n-2?
its a term 2 less than the infinity term (a sub n), hes just trying to show there are some terms there that are going to cancel with the last term.
The way I understanding this instantly but look at my professor like 🗿
How do you know which values will cancel out somewhere in the series?
just find a common pattern
i just checked that for the given term, what value of n does it meet it's additive identity, 1/(n+3) met its antimatter at n+2 which is not possible (sum is 1 to n), similarly, others require out-of-the-interval values of n for annihilation
He just an excellent math teacher. He give good lectures plus good exercises. I do love watching him
students next semester will be happy to have this video.
The man is happy.
On the problem at 11:00, I got the same answer but I didn't want to do the partial so I tried distributing the n first, then splitting the fraction. I ended up getting Sn= 1+ 1/n. Which ended up being 1 meaning it converged. Was what I did mathematically wrong?
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very good brain storming
I never thought my dumbass would understand calc 2 and yet here i am. If i can do it, anyone can i promise
You don't specify why you use and an A sub n-1 or an Asub n-2 term
You are absolute amazing I love this explanation!!!
Is it possible to know his name so that we can follow up with him
Tnx alot
But I didn't get the point @20:39 why is
(1/n+1- 1/n+3) considered as our last term
Ok I think I get it
It is because the given telescopic sequence is given as (1/n+1 - 1/n+3)
so we have to consider it as our last term right🤓
I was staring at my paper utterly dumfounded lol... thank you!
How do you know how many terms you need?
I think for a telescopic series only until som terms start to cancel
Sometimes you just stop where you feel like the canceling pattern is observed
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Last example was 🔥🔥
Great video! Thank you so much for making it.
Why can no one offer any explanation as to how you’ll know certain terms cancel out? I don’t see the pattern and I’d appreciate it if someone pointed it out before just telling me it is cancelling out eventually and taking your word for it.
@@Brad-qw1te Unless they failed I guess...
4:50
If n goes infinity, how did it becomes 0?
1/infinity is 0. If you divide 1 by a very large number, u are gonna get a very small number. In fact, as n approaches infinity (large number), result is gonna approach 0
well, thank you so muchy much 😍
does anyone know why at 14:07 he used -3?
bro has pink in his color set?
I understand when we put values in for a1,a2,a3...etc. But how and why do we start putting in, n-2, or n-1, An-5? is there a video that explains this process?
because we want to see if terms near the end will cancel out, that's why you do n-2 and n-1 to see if they have terms that cancel out too. Hence why its called a telescoping series cause it starts to collapse in on itself but you cannot automatically assume it cancels for every term. So he's showing that some terms are left uncancelled at the end
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Its seeming like the end terms are always 0 due to the limit of the function being 0. Is this true? Why do we have to look at the end behavior and not just the beginning terms?
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Thanks for letting me not going to math lab tomorrow. Or math lectures. Or literally anything math related besides the midterms and final
Why does the textbook like to write with i's? Is it necessary?
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my savior