For those wondering about the 2/(n^2) expression. Consider that the expanded terms he showed are constantly multiplying by something *smaller than one*; thus, the sum would constantly be getting smaller. If you remove these terms making your sum smaller, and just keep the final two terms, you will then have a series that is in fact larger than the initial series given. He also keeps the final two because together they make up a p-series, which is easy to tell divergence/convergence.
it seems to me like the last example would be much better solved by the ratio test, however this method explains the true fundamentals of the comparison test, so good job
thank you for all your help. ever since high school i've been watching you're videos and now I've passed my first year of math and i am finally FREE from it. thanks again!
i think in order for you to use the comparison test directly to show your series is div, the thing being compared must also be div (like example two) BUT it must also be a lower limit (not an upper limit like in patrickJMT's 2nd ex).
@mikegovikes it would be, but the point of this vid is to teach the limit and direct comparison tests. For alot of sums you can choose from almost any test to use on it, however what decides which test to use is basically a) preference b) ease of use c) and in some cases, like geometric series, the fact that you can actually define the sum of a series
These videos are so helpful, thanks so much. I have my final tomorrow and I've been trying to study the book, but having someone explain it makes it so much easier.
i agree, you can just look at it, realize it is close to being a p-series, and so say it diverges. you can say that only because of the limit comparison test (that is the justification); otherwise, you are just 'hand waving'
+Seandt Calculus -_- in the class room we just say omg hehehe but here 2 hours am done from all the chapter :3 i wish i've revised here to be the hero there :3
@floyd617 i leave those terms intact because it provides a useful expression: 2/n^2 and the series associated with that converges. you have to realize that you are trying to show the given expression is smaller than some convergent series (or larger than some divergent series). it is not mechanical to do this: how can one justify it? for this example, i found a way! (i hope i am remember correctly what is in the video)
An example where both the numerator and denominator are constants raised to "n" would be great! ^_^ thanks for the videos, it really helped me understand because i can rewind and watch again until i know what you're talking about.
@patrickJMT At the beginning, you said: [1 + sin(n)] / [10^n] is less than or equal to 2/10^n, "for all n is greater than or equal to 1". Why did you choose 1? Isn't it 0, because that's what our original series had n equal to? Thank you for your help! Your videos are great!
according to the p-series, 1/n^1/2 is divergent. since the limit of (sqrt(n)/n-1)/1/n^1/2 is 1, which is finite and >0, according to the limit comparison test, since the former is divergent, the latter is also divergent. you can also use divergence test since the limit of sqrt(n)/n-1 is not equal zero, it thus diverges.
think about what you said... n^(1/2) is the square root of n. consider n=4. 4^(1/2) = 2. 2*2 = 4. not 4^(1/4). Multiplying numbers with exponents is a slightly different rule set than multiplying fractions.
I think the issue is that video 122 in the Calc 2 playlist is a bit more of an intro to these ideas versus this video (121 in the playlist). So this one just uses a few terms such as p-series in a way that isn't immediately obvious.
Omg! I did not understand why only (2/n) and (1/n) were used until now! My math teacher showed this example a month ago and I remember something being negative ^^". Either my memory sucks or that explanation failed, but I wasn't the only one extremely confused that day.
why i didnt find yout videos earlier ??? whyyyy ? great lessons i am giving tommorow morning and now i am gonig to see as many of your videos as i can till 2mmorow
Because being less than a divergent series means nothing, but being less than a convergent series actual helps us. An, Bn = a sub n and b sub n; respectively For 0
I am a little confused on that part as well but my guess would be: n is a really big number. Even if you include the 3/n which would be 3/n*2/n*1/n= 6/n^3 this is still a convergent p series, so the answer comes out to be the same.
yeah, I did! The only thing that is challenging for me in Calc is the intermediate Algebra. I would rather screw up on the simple stuff than not have the intellectual capacity for the fun stuff. I hope that is not to complicated for you?!
@AznVamp24 if you use the direct comparison test and show that the sum of the series is smaller than the sum of the series associated with 1/n you have not shown anything as that latter series diverges.
I hope you make a decent living off of these videos and the apps (I bought many of them, myself!). They are so unbelievably helpful and have saved my grade on many occasions! Thank you :D
@pochankitty not free. dont you see the commercials? lol. great teacher nonetheless. definitely always rely on you, patrickJMT, when my teachers fail to teach
@stealinglemons I do not think he covered P-Series. I had to look it up in my book and get a quick definition of it. P series is refering to the exponent of the 1 / n^p. If p > 1 ; Converges if p
that's just because infinity isn't a number at all. It's just a concept used to understand that there is no largest number. 1^anything is 1 though... That's really the point of using limits if you're going to use infinity. go to wolframalpha and put this in: lim as n approaches infinity 1^n
Sir, could you please answer me these questions : 1) You figure out the bsubn series by figuring out the highest powers in both the numerator and the denominator? 2) Suppose i'm given a question and asked to find whether it converges or diverges, can i just ignore the direct comparison and do the limit comparison test? cuz the limit test seems easier to me than the direct one. 3) for a p series, can we conclude that the series diverges if p< 1? Thanks in advance.
I finally understood last example for example we have 0,05 0,04 0,03 0,02 0,01 multiply first three numbers 0,05*0,04*0,03=0,00006 multiply last two numbers 0,02*0,01=0,0002 see 0,00006
not sure but maybe the reason for why he uses the last two terms [2/n and 1/n] for the second problem is because n!/n^n maybe > at a certain point then 1/n
Yeah, a lot of people are confused about the last example where you compare the original to (2/n)(1/n). What's stopping you from just comparing it to only (1/n)? And why are those valid choices since they're both less than 1 as well? Thank you! :]
+samrath singh I tried the problem using the ratio test and got a solid answer. When you use the LCT or DCT, then this becomes a shaky argument. I confirmed with my professor, and he said this is a ratio test argument. Btw I did get convergent after using the ratio test. Hope this helps!
No, if you took (1/n), then the comparison test would be inconclusive. Since (1/n) > (n! / n^n) , if Sum(1/n) diverges then you CANNOT conclude anything about (n! / n^n). In which case, you need to then pick a different sequence to compare to. When you pick (2/n^2), btw you couldve picked any constant divided by n^2, since (2/n^2) > (n! / n^n) AND (2/n^2) converges, then (n! / n^n) MUST also converge. Picking the harmonic series leaves with you an inconclusive argument. This happens often with the comparison test.
Also, 1 other question. How did you conclude the 2 / n^2 in the next problem? I see how you got it, but why are you able to do that? Is there another (better) series test to tackle this problem with? It's just hard for me to wrap my head around the 2 / n^2 part.
-_- Don't say he is wrong... When you multiply you add powers, you don't multiply them. so your problem would be 1/2+1/2 which does indeed equal 1, so it does equal n. To make it equal to n^(1/4) you would need a power to another power (that is when you multiply powers. So, [n^(1/2)]^(1/2) would equal n^(1/4)... Don't mess with JMT man
your video is so helpful for me to understand the concept！But may i want to know is there any differences between comparison and limit comparison？I cannot find any difference in the three examples
I understand how sin(n) is always -1 < sin(n) < 1 .... so I thought it turned into (-1)^n. How do you simply come up with 2 / 10^n as your b-sub-n? Would you mind being at least slightly detailed? I'm just a little confused. Thank you in advance. I hope you have the time to respond. Your videos are excellent and I am extremely appreciative of the work you do.
what made you pick out the last two instead of the last one? because if you only used 1/n, it would be divergent because its harmonic series but you chose the last two making it a p-series?
This I can actually answer, the numerator 1+sin(n) can never exceed 2 because sin(n) wavers between -1 and 1. Therefore 2/10^n will always be greater. To be more clear when I did this problem I used 3/10^n since its more obvious that it'll always be larger.
I am confused as to why you left out the 3/n but not the 2/n and 1/n ? I see how the terms you left out are less than one, but why not 2/n and 1/n? Maybe this is a silly question, but I'm not seeing it for some reason
@patrickJMT So are you saying that I could have used the last THREE terms as well? That would have still proved the exact same thing, no? btw, YOU ARE AWESOME THANK YOU SOO MUCH!!!
why did u use limit comparison test in the second one instead of direct comparison... i get that an is bigger than bn, and because of the p series it was already divergent, i could have used direct comparison ... ?
when do you know to use limit comparison or direct comparison? do you just use limit comparison when you cant do direct? would you be able to do limit comparison instead of direct and get the same answer? lets say if i did limit comparison for your first example. THANKS ALOT!
For the problem at 3:13, why don`t you say that the series is divergent if its bigger than the divergent p-series. Why use the limit comparison test instead?
wait, in the last example, why do you keep the last two terms: 2/n and 1/n ? i mean, how do you know not to just keep the last term 1/n? (which would then make it divergent)
I understand we compared it to 1/n which dv. Im just a little confused about all the test and what it needs to be greater or less then in order for it to cv or dv
For those wondering about the 2/(n^2) expression.
Consider that the expanded terms he showed are constantly multiplying by something *smaller than one*; thus, the sum would constantly be getting smaller. If you remove these terms making your sum smaller, and just keep the final two terms, you will then have a series that is in fact larger than the initial series given.
He also keeps the final two because together they make up a p-series, which is easy to tell divergence/convergence.
it seems to me like the last example would be much better solved by the ratio test, however this method explains the true fundamentals of the comparison test, so good job
it's so refreshing seeing a fellow left-hander teach math. this was a very helpful video. thanks.
Once again I get stuck on a problem and magically ur video is on the same one! Thanks sir
much better than my teachers lessons in class! I can use your videos with the CK12 calc book to make my understanding much better, thanks :)
thank you for all your help. ever since high school i've been watching you're videos and now I've passed my first year of math and i am finally FREE from it. thanks again!
i think in order for you to use the comparison test directly to show your series is div, the thing being compared must also be div (like example two) BUT it must also be a lower limit (not an upper limit like in patrickJMT's 2nd ex).
When multiplying the same base with different powers, add the powers.
IE: 2^(1/4)*2^(3/4)=2
Your videos are the only ones that can actually seem to make some sense out of this for me. Thank you thank you thank you for making them!
@mikegovikes it would be, but the point of this vid is to teach the limit and direct comparison tests. For alot of sums you can choose from almost any test to use on it, however what decides which test to use is basically
a) preference
b) ease of use
c) and in some cases, like geometric series, the fact that you can actually define the sum of a series
I really enjoy your calm voice
patrickJMT is the god of calculus thank you so much for the help
These videos are so helpful, thanks so much. I have my final tomorrow and I've been trying to study the book, but having someone explain it makes it so much easier.
Thank you so very much. My calculus teacher is out of control. Thank you so much.
You are so much more clear than my teacher, thank you!
i agree, you can just look at it, realize it is close to being a p-series, and so say it diverges. you can say that only because of the limit comparison test (that is the justification); otherwise, you are just 'hand waving'
Why do "professors" make this so confusing?
In awe of a Brilliant Mind
+Seandt Calculus -_- in the class room we just say omg hehehe but here 2 hours am done from all the chapter :3 i wish i've revised here to be the hero there :3
ty patrickjmt
you just saved me 1 hour of reading through the difficult language of my calc book.
@floyd617 i leave those terms intact because it provides a useful expression: 2/n^2 and the series associated with that converges. you have to realize that you are trying to show the given expression is smaller than some convergent series (or larger than some divergent series). it is not mechanical to do this: how can one justify it? for this example, i found a way!
(i hope i am remember correctly what is in the video)
An example where both the numerator and denominator are constants raised to "n" would be great! ^_^
thanks for the videos, it really helped me understand because i can rewind and watch again until i know what you're talking about.
that third one though. haha I don't know how I would've thought of that on my own
+kalef1234 3rd u need to use lembert directly while us ee n! and a^n
For that problem, I think it would have been easier to use the ratio test
ratio test is really useful for factorials.
andy brisman I tested it, the ratio test is inconclusive for that problem
@@carl7018 I have reached to (n/n+1)^(n+1) then putting infinity causes to inconclusion. Comparation test is the only way i think.
@patrickJMT At the beginning, you said:
[1 + sin(n)] / [10^n] is less than or equal to 2/10^n, "for all n is greater than or equal to 1". Why did you choose 1? Isn't it 0, because that's what our original series had n equal to?
Thank you for your help! Your videos are great!
according to the p-series, 1/n^1/2 is divergent. since the limit of (sqrt(n)/n-1)/1/n^1/2 is 1, which is finite and >0, according to the limit comparison test, since the former is divergent, the latter is also divergent. you can also use divergence test since the limit of sqrt(n)/n-1 is not equal zero, it thus diverges.
think about what you said... n^(1/2) is the square root of n. consider n=4. 4^(1/2) = 2.
2*2 = 4. not 4^(1/4).
Multiplying numbers with exponents is a slightly different rule set than multiplying fractions.
I think the issue is that video 122 in the Calc 2 playlist is a bit more of an intro to these ideas versus this video (121 in the playlist). So this one just uses a few terms such as p-series in a way that isn't immediately obvious.
omg you explain these things better than my professor. your tutorials are awesome! :)
Thattutorguy is freaking overrunning every math tutor channel.....
Omg! I did not understand why only (2/n) and (1/n) were used until now! My math teacher showed this example a month ago and I remember something being negative ^^". Either my memory sucks or that explanation failed, but I wasn't the only one extremely confused that day.
why i didnt find yout videos earlier ??? whyyyy ?
great lessons i am giving tommorow morning and now i am gonig to see as many of your videos as i can till 2mmorow
Because being less than a divergent series means nothing, but being less than a convergent series actual helps us.
An, Bn = a sub n and b sub n; respectively
For 0
I am a little confused on that part as well but my guess would be: n is a really big number. Even if you include the 3/n which would be 3/n*2/n*1/n= 6/n^3 this is still a convergent p series, so the answer comes out to be the same.
yeah, I did! The only thing that is challenging for me in Calc is the intermediate Algebra. I would rather screw up on the simple stuff than not have the intellectual capacity for the fun stuff. I hope that is not to complicated for you?!
@AznVamp24 if you use the direct comparison test and show that the sum of the series is smaller than the sum of the series associated with 1/n you have not shown anything as that latter series diverges.
I hope you make a decent living off of these videos and the apps (I bought many of them, myself!). They are so unbelievably helpful and have saved my grade on many occasions! Thank you :D
@pochankitty not free. dont you see the commercials? lol. great teacher nonetheless. definitely always rely on you, patrickJMT, when my teachers fail to teach
FXXX YEA. IAM going to pass my CACL Test, THANKS MAN!
@Payme4ril p-series
1/n^c
with c being a constant. if c > 1 it converges
if c
YOU ARE THE BEST !!
@stealinglemons I do not think he covered P-Series. I had to look it up in my book and get a quick definition of it.
P series is refering to the exponent of the 1 / n^p. If p > 1 ; Converges if p
(n^(1/2))/n^1 = 1/n^(1/2) ; u are just subtracting exponents
Good eye there xSilver.....
I have to agree with my math theacher... I'm not smart, just stubborn.
I love how you're trying not to ruin the surprise of the second example. Chortle.
that's just because infinity isn't a number at all. It's just a concept used to understand that there is no largest number. 1^anything is 1 though... That's really the point of using limits if you're going to use infinity. go to wolframalpha
and put this in: lim as n approaches infinity 1^n
it'd also work since 6/n^3 is also a convergent p-series
YOu rock! thanks so much for the help, I'm studying civil engineering right now and this is really good stuff because it helps.
Again, thanks so much.
Sir, could you please answer me these questions :
1) You figure out the bsubn series by figuring out the highest powers in both the numerator and the denominator?
2) Suppose i'm given a question and asked to find whether it converges or diverges, can i just ignore the direct comparison and do the limit comparison test? cuz the limit test seems easier to me than the direct one.
3) for a p series, can we conclude that the series diverges if p< 1?
Thanks in advance.
the second problem you did, wouldn’t it be easier if you just said that since an>bn and bn is divergent, then an will also be divergent?
I finally understood last example
for example we have 0,05 0,04 0,03 0,02 0,01
multiply first three numbers 0,05*0,04*0,03=0,00006
multiply last two numbers 0,02*0,01=0,0002
see 0,00006
thank you so much, this is much more helpful than lecture
@AznVamp24 1/n is a harmonic series which is divergent...but 2/n^2 is convergent
not sure but maybe the reason for why he uses the last two terms [2/n and 1/n] for the second problem is because n!/n^n maybe > at a certain point then 1/n
Yeah, a lot of people are confused about the last example where you compare the original to (2/n)(1/n).
What's stopping you from just comparing it to only (1/n)? And why are those valid choices since they're both less than 1 as well?
Thank you! :]
In third problem why you took 2 terms, if we only take last 1/n then it would be harmonic and divergent ?
+samrath singh I tried the problem using the ratio test and got a solid answer. When you use the LCT or DCT, then this becomes a shaky argument. I confirmed with my professor, and he said this is a ratio test argument. Btw I did get convergent after using the ratio test. Hope this helps!
No, if you took (1/n), then the comparison test would be inconclusive. Since (1/n) > (n! / n^n) , if Sum(1/n) diverges then you CANNOT conclude anything about (n! / n^n). In which case, you need to then pick a different sequence to compare to. When you pick (2/n^2), btw you couldve picked any constant divided by n^2, since (2/n^2) > (n! / n^n) AND (2/n^2) converges, then (n! / n^n) MUST also converge. Picking the harmonic series leaves with you an inconclusive argument. This happens often with the comparison test.
Also, 1 other question. How did you conclude the 2 / n^2 in the next problem? I see how you got it, but why are you able to do that? Is there another (better) series test to tackle this problem with? It's just hard for me to wrap my head around the 2 / n^2 part.
-_- Don't say he is wrong... When you multiply you add powers, you don't multiply them. so your problem would be 1/2+1/2 which does indeed equal 1, so it does equal n. To make it equal to n^(1/4) you would need a power to another power (that is when you multiply powers. So, [n^(1/2)]^(1/2) would equal n^(1/4)... Don't mess with JMT man
your video is so helpful for me to understand the concept！But may i want to know is there any differences between comparison and limit comparison？I cannot find any difference in the three examples
have fun :)
yeah he does
Very helpful. Keep it up, I got a calc test coming up :P
I understand how sin(n) is always -1 < sin(n) < 1 .... so I thought it turned into (-1)^n. How do you simply come up with 2 / 10^n as your b-sub-n? Would you mind being at least slightly detailed? I'm just a little confused. Thank you in advance. I hope you have the time to respond. Your videos are excellent and I am extremely appreciative of the work you do.
you are da best. you should do videos on physics too!
this is one of the hardestest part of series to grasp imo haha
what made you pick out the last two instead of the last one? because if you only used 1/n, it would be divergent because its harmonic series but you chose the last two making it a p-series?
I think I get it now. You have to test An for the 2nd example because if the limit of An/Bn happened to be 0 it would converge?
n! (called n factorial) is defined as 1*2*3*...*(n-1)*n
@eternity23211 part of the fun is figuring that out
the original series is not a p-series though
Not sure what we are looking for when doing either comparison test? Something that looks like a geometric or p-series?
do you have a video that explains the concept of the direct comparison and integral comparison tests, without just jumping straight into examples?
it's about time he got some commercials
on the second example, when you kept the last two terms, why did you keep only the last two?
but getting paid like i am 10 years behind it :)
@patrickJMT So are you saying that I could have used the last THREE terms as well? That would have still proved the exact same thing, no?
For example 1: Why is (1+sin(n)/10^n) < 2/10^n for all n>1 since the limit of the problem says n=0 ?
Kevin Park yes I am also too confused? ?
This I can actually answer, the numerator 1+sin(n) can never exceed 2 because sin(n) wavers between -1 and 1. Therefore 2/10^n will always be greater. To be more clear when I did this problem I used 3/10^n since its more obvious that it'll always be larger.
I am confused as to why you left out the 3/n but not the 2/n and 1/n ? I see how the terms you left out are less than one, but why not 2/n and 1/n? Maybe this is a silly question, but I'm not seeing it for some reason
@patrickJMT So are you saying that I could have used the last THREE terms as well? That would have still proved the exact same thing, no? btw, YOU ARE AWESOME THANK YOU SOO MUCH!!!
great video.
Ratio test on the last one would have worked as well
and what if the limit doesn't give out a finite number? does the test fail and you just stop?
couldnt we have used the direct comparison again for the second one and avoiding doing limits?
why did u use limit comparison test in the second one instead of direct comparison... i get that an is bigger than bn, and because of the p series it was already divergent, i could have used direct comparison ... ?
when do you know to use limit comparison or direct comparison? do you just use limit comparison when you cant do direct?
would you be able to do limit comparison instead of direct and get the same answer? lets say if i did limit comparison for your first example.
THANKS ALOT!
@patrickJMT hahaha thats so evil. Thanks for making this easier buddy
For the problem at 3:13, why don`t you say that the series is divergent if its bigger than the divergent p-series. Why use the limit comparison test instead?
on example 1, why does 1/10 have to be less than 1? shouldn't it be 1/(10^n) less than 1/(n^2)?
@asianconfusion many years of making videos through many moods and levels of health.
@patrickJMT thanks
wait, in the last example, why do you keep the last two terms:
2/n and 1/n ?
i mean, how do you know not to just keep the last term 1/n? (which would then make it divergent)
I don't understand why the second example is divergent. I thought that 1/n
^1/2 would converge to zero as n approaches infinity??
I wish you were my math teacher.
Why can we not use Divergence test?
Thx u Patrickjmt
For the first problem, why does n have to be greater than or equal to 1?
You said sinx is between -1 and 1 so what is cosx in between?
Is the direct comparison test the same as the basic comparison test
Doesn't the limit comparison only work for positive term series?
if you wanted to, could you use (3/n)? or is it only (2/n &1/n)?
hey Patrick why did we use LCT on the second example. why didn't we use comparison test. I'm confused when do we use LCT and Whe do we use CT.
I understand we compared it to 1/n which dv. Im just a little confused about all the test and what it needs to be greater or less then in order for it to cv or dv
thank you!