Telescoping Series , Finding the Sum, Example 1
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- เผยแพร่เมื่อ 29 ก.ย. 2024
- Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! Telescoping Series , Finding the Sum, Example 1. Here I find a formula for a series that is telescoping, use partial fractions to decompose the formula, look at partial sums, and take a limit to find the sum.
You do math in permanent marker. That confidence is admirable, sir.
10/10
😎😎😉
A god doesn’t make mistakes
i do all my math in pen. even exams. always have since i was a kid. I'm not super confident though rofl
Shut up indian
Thank you for this. For the life of me I could not understand this concept by reading about it in the calculus text book. You really cleared it up for me... Thank You!
It's rare to find a video without a *thumbs down*. Amazing walk-through.
I was lost. I watched the video and I saw the light!!!
this video actually just saved me from a total breakdown (i only cried for a teeny but lol) thank you so much this makes perfect sense now
thanks for recommending and congrats on doing so well! :)
You're a lifesaver! Better than my prof to say the least...
I thought I was lost at the the last part with k, but I forgot to multiply the result with the numerator I factored out. Awesome video man, thanks for the help.
I know im not suppose to do this but I dont even look at lecture slides anymore when reviewing for an exam at the last week, I would just go to your channel. thanks alot senpai! :)
Great Job! I understood everything.
Please keep doing what you're doing
You are an amazing teacher
Thanks! I am preparing for JEE Advanced 2022 India....and this helped me so much!
How it went?
I think if you keep going to n=5, n=6, etc, that -1/5 and -1/6 actually do cancel out, so you can ignore that part and just look at the 1+(1/2) part.
Thank you for not doing n/n(n+1) like everyone else!
Excellent! You saved my life, thank you!
This channel might have just saved my finals :')
Thx u PatrickJMT
How do you find the sum for a problem that doesn’t have stuff in the denominator
This helped a lot, thanks!
THANK YOU!! So helpful!
you write your sigmas like a math gangsta! thanks :)
how do we detect a telescoping serie
take the sums... go out a few until you see that terms are repeating but with opposite signs(positive 2/3 and then a few sums later negative 2/3). If you have something like that where you can cancel terms then you have telescoping series.
watched the video 10 times before i finally understood it
At 2:44 why can't we put the series to each term and then say that series 1/n is divergent (harmonic).
thanks patrick
Omg I finally understand this thank you i luv u
This video was great up until you wrote the formula for partial sums (4:20 - 5:05). The first two terms: 1 + 1/2 makes sense, but it's not clear how you obtained the last two terms: 1/k+1 - 1/k+2.
Isn't (1/n) divergent by p-series test. Thus the entire sum is divergent?
no.. If the n>0 then the series converges. n in this case is 1, which is greater than zero
Gabriel Zanandrea this is a year later, but I hope you know you are wrong lol. p>1 converges. Anything else is a harmonic series
I love you Patrick! Bae for life
This diverges right?
Yes because 15/4 is greater then -1to1
I still don't get how you expanded and got the last two expressions :((
did you get it now?
Love ya.
do we always do the first 4 terms?
No, you just do as many terms as it takes to see a pattern. Sometimes it takes more, sometimes it takes less. In this case, he could've technically seen the pattern after only three terms, and in others you might not find a pattern, so you record a few more to see what happens.
nice
ahhh I see, thanks.
I dont understand anything your doing
cc
who know how to cuff me esperrys
Can you take my midterm for me tomorrow?
+Akhil Thuthika seems like we all have mid terms tommorow...
RIP
we all do...
it's been two years how was the midterm... lol
I have a feeling he didn't pass
Izco Mdz lol
that sigma looks dope af
LOL!
well yeah
i left the WebAssign guy for you.
i am not sure what that means :)
WebAssign is an online math program that profs use to assign hw and stuff. it has "help videos" done by a guy I've named Jerry but I've left him to subscribe to your channel :)
Webassign sucks
Jerry is sobbing rn in the darkness: "Hello darkness my old friend"
@@patrickjmt I could not even find a video for the first problem involving this process in Webassign. It is mentioned in the text but the terms that needed to be cancelled showed up more quickly ... I had no idea that we should continue taking sums...
(they just did an interface update, summer 2019 btw)
Professor did not even cover telescoping series examples in class although it was included in online hw. We have only gone over the eight(8) different tests.
I loove this video. Thank you for the step by step. I can start my hw now.
Man you are a legend. We have learned so many things from you. Whenever i get stuck while solving my notes, i check your channel and you already have the videos of those topics being explained briefly. Great job (Y)
thanks!
For those asking about the 1/k+1 thing: that's correct. He found that the partial sum when n=4 (because he used the first four terms) was [1 + 1/2 - 1/5 - 1/6]. Since this is the sum when n = k = 4, the 1/5 is rewritten as 1/k+1. Think about if he just did the first three terms, so that n = k = 3. You'd be left with [1 + 1/2 - 1/4 - 1/5]. The 1/4 is now 1/k+1 and the 1/5 is 1/k+2.
Hope I helped.
Yooo thanks fro clarifying that !
1/5,1/6 terms were unnecessary as they were going to be 0 anyway..
may your crops grow and your skin shine
Sometimes I wonder why my my math teacher didn’t explain it like this. He just said to write out some terms towards the end and beginning of the series and just figure out what cancels. This is much quicker. And thanks for the explanation of the k+1 thing!
im from 2021 and i want to thank you for clarifying this part
no clue where the K+1 and K+2 comes from, iv rewinded it like 10 times...
It's cause he ended n=4 and had 1 + (1/2) -(1/5) - (1/6) when he ended at n=4. the (1/5) term is (1/n+1) and the (1/6) term is 1/(n+2)
Cameron Aminian thanks man, yea i finally got it after practicing the problem a few more times lol
@@cameronaminian9442 hi..can i know at least what n term we must write?He ended at n=4..if i ended n=3 can?
Man, you a bold man for using a sharpie! I would need so much paper if I did my homework in pen. XD
He's doing an example problem
lol I still do.....
and should be putting stock in white out now that we're in series. 😂
this video will be saving people who didnt pay attention in class even after 10 years
I know I'm about a decade late, but this video was a lifesaver! I'm working on telescoping sums for my discrete mathematics course, and I didn't get how to break apart the denominator until this video. Genuinely thank you so much for making this content public!
this thing is gona fuckin drive me crazy, omg!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
:S
@mlleJMHRG very happy to help :)
Why. Why. Why do I need to do any of this absurdity to learn how to program a computer. Math instruction in this country is a fucking farce.
So it is not clear on EXACTLY how you got the last two terms of the sequence 1 over k+1 and 1over K +2
you are welcome :)
This reaffirms my hatred for calculus
Man, 3 years ago, when I took Calculus 2, I used this video to study telescoping series. 3 years later, in an upper division math class, in order to refresh my memory for hw, I am using this video again. Thank you patrickJMT !!!
( you are welcome ) x 100 :)
Cal 2 final in 13 minutes and I have just learned something very important. Thank you sir!
@joeratti ha, sweet!
Why can't we take the limit right away? (I know taking it will yield the wrong answer, but I don't understand why.)
i totally wouldve just left 3/2 as my answer conpletely forgetting the 5/2. This guy's either a beast or does multiple takes till perfection hahaha
I don't know how many times I can say thank you. You've gotten me through five math classes and working on two more, at both high school and university levels. My lifesaver!
k? wtf man u know it's always infinity.
isnt the series divergent since the sum of [1/n + 1/(n+2)] is infinite? Also, the lim you took that gave you 3/2 proves it is divergent since it would equal 0 if it was possible for it to converge.
finallyyyyy a sum that doesnt have x/n - (y/n+1) !!!! been all over the internet trying to find examples thank you!!
nth div test, goemetric, p series, alternating, integral, and ratio/root test are all under my control; but telescoping and comparison tests bother me to no end! Telescoping is a pain in the ass for obvious reasons, but the thing that gets me with comparison test is choosing the right Bn to compare your An to : (
can you be my college calc prof??
maybe!
Your videos are life savers man. Keep doing what you're doing! You're the best!
Missed a 50 minute class. Made up for it in 15 minute in the comfort of my room. Keep up the good work man.
I believe the (1/k+1) is incorrect. The n in that case is 3, and 3 +1 = 4, not 5. With that being sad, i believe the pattern is only the (1/k+2). Correct me if Im wrong
You wrong. It is not about the denominator it is about the pattern. We cannot have 1/k+2 without having 1/k+1 first
This is where you get it wrong. And by the way if the pattern was [1/n - 1/n+3] you should continue until 1/k+3
The sum or limit is actually 1 + 1/2 = 3/2, because for any value of 'n' there is nothing left or the rest equals 0. For instance, n = 5 -> 1/5 - 1/5+2 = 1/5 - 1/7 ; n = 6 -> 1/6 - 1/8 ; n = 7 -> 1/7 - 1/9... There are all possible values for (1/n) and its opposite - (1/n), then the rest is 0.
Thanks alot really helpfull
so use decomposition to rewrite, take partial sum, writing out the terms for a bit to see a pattern, then write generic involving k, then take limit of series from n to k of the generic thing we found from writing out terms. that's it (plus multiplying by coefficient we took out earlier) Hope I got it! And this was just to evaluate the series...is that the same as finding the sum?
So bad he can’t show the straight problem .. all back and forth ..
will it be possible to pass exams with flying colors by simply relying on your videos, because i understand you best
mbuso mabuza yes
Ok, I'm very sleepy and not wearing glasses, but I honestly thought the title was "Telescoping Series (like astronomical telescope), Finding the Sun".
Bro this refresher, thank you so much man
Thank you so much! I've been trying to figure out how to do this on my own since it's going to be on my exam but we didn't actually cover it in the classes (two teachers, 1 has lectures, the other one makes exams...makes sense). I felt so dumb for not understanding it (the numbers at the end just seemed random, I didn't realize it's because the rest cancels itself out) and I couldn't find it anywhere. Now I'm actually kind of angry that it's soooo easy and I was stuck on it for hours :D You're a lifesaver. Well, hopefully. I haven't taken the exam yet :D
This method is not only far longer, but it tends to 0, not 15/4
Denominator tends to infinity while numerator is at 5. Therefore, the function gets infinitesimal closer to 0
I got 5/[(n^2)-1] as my equation starting from n=2 to infinity which works, can someone explain how i'd find the first term since I need it to find the sum?
I don
t understand something. Why did you use 1/(k+1)? Wasn't there any other way?
i am asking the same question ...
So on example A, when doing the Sum with the sum to a K constant, why wouldnt the other terms work the same way? Such as the 1 being converted to n-3....since we stopped at n=4. and the 1/2 turn into 1/(n-2)?
Thanks. I was searching for this kind of video😊
Thanks a lot
shouldn't it be 1/k+2 at 4:51?
Shawn Isi *wonders the same thing too
with all of the telescopic problems i've worked, the k's never play a part because they always get canceled out when you take the limit.
Thank you sir for this imformative video its made sum easier for me
Excellent. Just Excellent. Thank you Patrick
3:36 is the important part:
Once you do the first couple of indexes, you find the pattern as far as which terms cancel out.
at 4:48, im confused about why -1/5 and -1/6 stay. Wouldnt they simply be cancelled out when we reach n=5 and=6 ?
+Amir Haghi they do go away, but he was using those terms to find the nth term
He did the sum from 1 to k, and decided to stop at the 4th term, making k = 4. If you plug a 4 into that equation, you get [1 +1/2 - 1/5 - 1/6], which is the answer that he got. If you added the first 10 terms, you could just plug a 10 into that equation for k, and that would be the answer. So yes, they cancel after you go past k, but he was finding the sum at k.
He did say that it will cancel out as k goes to infinity because it will have k go to 0
correct me if i'm wrong , the reason -1/5 = -1/(k+1) is because you expand till n=4 ? thanks
patrick, what do we do if only 1 term is cancelling or nothing is cancelling when you do the sum
This is a very complete explaination,thanks to u 🖒..now i understand the concept 😁
Your videos are really helpful and I always use them to review before a major test, thanks so much! One request, please make the videos a lot louder.
It's been a while since I done these and thus I completely forgot the method. This video helped a ton.
Students need a great teachers like you. Thank you for making our life simple
Was just studying for a test. You helped a lot. Thank you.
Thank you for explaining this so well, my professor didn't make sense
the only thing better than patrickJMT is patrickJMT in HD :D
did you get a new camera? The video quality is a lot nicer :)
I realize this comment is pretty old but for the people out there who may be confused too, the k is usually denoted as n in a textbook so instead you may see n+1 and n+2. (-1/5) becomes (-1/k+1) because the last term he solved for was the 4th term. How does this term number (4 or n, or in this case k) become 5? By adding 1 to it. Hence the denominator becoming k+1 or n+1. Same for the next one. (-1/6) must become a generic expression including n. n is 4 so again how does it become 6? by adding 2
My teacher wrote a similar thing but with " (n+a)(n-b)". How I my suppose to prove that it converge? I mean, I can do a example like in this video, but it's a example. That doesn't prove that it works for every "a" and "b" if "a" and "b" are some integers.