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.
@@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
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.
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
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 RUclips TV with the Organic Chemistry Tutor.
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.
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
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
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?
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
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.
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
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)].
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
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?
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.
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
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.
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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
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
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 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?
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
i dont like south africa ooga booga people
@@gartyqam What about you, are you a king/ queen?
@@04agnishikhamajumder24 white
@@gartyqam don't every buy a tesla, use paypal, or go to mars then
He just an excellent math teacher. He give good lectures plus good exercises. I do love watching him
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 RUclips TV with the Organic Chemistry Tutor.
How beautifully this man has explained so complex thing. Respect❤❤
You deserve an award . Seriously, you are the best in all your videos
series are literally going to be the death of me. the rest of calc 2 has been so easy up until now
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?
@@Jee2024IIT huh?
@@cameroneinstein546 nothing
Removing the red point left in 4:10 is so SATISFYIIING!
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.
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!
Thank you! This helped me with a tricky problem on my homework!
Love you💚. You just explained my whole Calculus semester
people who teaches on black screens are gods
It’s much better for eyes too.
There is one God. The God of the Bible.
Last example was 🔥🔥
لولاك ضعنا ياخي 😔❤️❤️❤️❤️❤️
bro youre a life saver i swear
what an amazing instructor
bless your heart! you simplified this for me in the best way possible!
3 years later and you've helped me big time
i am so greatful for this thing u have taught all of us it was so helping for me i am so greatful.thnks
YOU, SIR, ARE MY SAVIOR! THANK YOU SO MUCH!
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
I hope you never quit youtube
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?
Thank you for explaining everything in a very simple way!
at 2:28 why did you use n-2 not n-1 or something else?
You deserve Nobel Peace Prize for students✌️.
You deserve to be a professor actually teaching in college. My calc 2 professor barely helps me until i understand more and more
Buddy you are more than god for me, as my guru (teachers)💌
You are absolute amazing I love this explanation!!!
You are indeed a savior
omg perfect timing, we're learning this in my calculus class now
same
Really brother , you explained so well , such that u cleared my doubt in once❤❤❤❤❤❤❤❤
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
you are soo...ooo good at explaining stuff.
ur single handedly responsible for my degrees ty i owe u my first born
you are better than my university's professor
do you know that in my country we study lessons of college to inter the college :) and yeah you are saving me now ...thank you a lot
انتي عربية بس من اي دولة؟
@@ousali1340 morocco 😊
You're a savior.
I met you when your channel had at least 300K subs, you re reaching 1 M go ahead man
So helpful sir. I can say after this video, I get it.
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
I was staring at my paper utterly dumfounded lol... thank you!
This is most likely the easiest method for determining what the series converges to. You just have to be good at partial fractions
IM SO GLAD YOU EXIST
Was a little confused. Thank you so much for a well structured explanation!
Thank you sooo much all your videos help a lot!!!
Thanks for letting me not going to math lab tomorrow. Or math lectures. Or literally anything math related besides the midterms and final
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
EXCEPTIONALLY GOOD
Simple steps, nice...
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.
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)].
Great video! Thank you so much for making it.
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
Can someone please tell how do we get to know when to use an -1 ,an-2 or an-3
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.
Where is that video i need the link of that you told in this video
Sir can i have some request?? I am strugling right now in writing a series in sigma notation... can you please make a video on it???:) thanks a lot
Very helpful and informative! Thank you!
very good brain storming
You are the best.. thank you very much
Great video MAN
Why does the textbook like to write with i's? Is it necessary?
the last example is only possible if we assume an-7 and an-6... can this only be proven if we write out all 7 terms ?
You explain way better then my 'professors' at uni.
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?
are telescoping and difference series same thing in sequence and series
13:44 Wait why does (n+1) goes first then (n+3) shouldn't they be switched?
Thank you for this video. It helps a lot!
Thank u very much sir
Thanks...you saved me
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
for the first problem, wouldn't the 1/n+2 technically cancel if you were to add one more sequence?
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.
thank you so much.
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
Huh?
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🤓
awesome video
you are the best! Thank you so much.
I never thought my dumbass would understand calc 2 and yet here i am. If i can do it, anyone can i promise
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.
Thank you so much
Thank you sooooo much you are amazing!
thank you u r the best
sir in the question two they have given n=1 to infinity then why have you taken n=-1 to find the value of b
Because partial fraction decomposition has nothing to do with the sum and holds for any value of n.
The way I understanding this instantly but look at my professor like 🗿
Good job
THANK YOU SO SO MUCH!
You're too good😊😊😊 thanks im happy more videos please 😊
Happy to say that I understand this theorm, but some of the other theorms from your other videos... Confusing a bit
You are hero ❤😭