Limit Comparison Test and Direct Comparison Test

[Thangiee12]

what would you compare 1/n! to?

[kikoisme1]

quick question??? what if you are dealing with Ln's for example lnN/n^(3/2) also for the second example couldn't you have just used the direct comparison because it was 1/n^(1/2) p=1/2<1 which makes the series divergent

[stickfigure578]

You are so much more clear than my teacher, thank you!

[khilozozo02]

I don't get it...

[patrickJMT]

@khilozozo02 what dont you get

[RaulTheSunshineMan]

Nice.

[CaptainAngelos]

Was there ever a video on p-series?

[avpescatore]

it's about time he got some commercials

[melisantika]

I am very happy to see the vidoe Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges from you, hopefully the others also are happy for You

[Onepissite]

I am very happy to see the vidoe Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges. after you give this

[Ondelendo]

I Love The Video It Can Increase My Knowledge Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges

[bebeheuy]

Steady I Really Like This Video Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges

[willamricard]

Nice Video That You Share , So Very Nice Thanks You Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges.

[imegatrone]

I Really Like The Video From Your Limit Comparison Test and Direct Comparison Test - Using the Limit Comparison and Direct Comparison Test to Determine if a Series Converges or Diverges

[VaCHillBand]

Thank you Patrick.

[umbragami]

@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!

[musicpharoah]

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.

[dsmTooki]

Once again I get stuck on a problem and magically ur video is on the same one! Thanks sir

[pepteamsergi09]

Nice



[elytheoneandonly]

i think the ratio test works better for the last one.

[mahisona]

Thanks a lot for the very 1st problem of this video. I had the same one in my Hw.:D

[Fg5x3]

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.

[eternity23211]

how do we know when to use the direct comparison test or the limit comparison test? im confused

[patrickJMT]

@eternity23211 part of the fun is figuring that out

[ian559fresno]

@patrickJMT hahaha thats so evil. Thanks for making this easier buddy

[ShrodingersPerson]

Nevermind I figured it out. :)

[ShrodingersPerson]

So if the ratio at the end happened to be -1 or less than zero then what happens? What does it mean for An?

[DarthYoungling]

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? 

[travdoeshave88]

for 1/(n!) could i say 2/(n^2)is > or = to 1/(n!) or is my math just plain wrong

1/(n!): 1/1+1/21/6+1/24... 1/[(n-1)!]+1/(n!)

2/(n^2):2+1/2+2/9+1/8... 2/[(n-1)^2]+2/(n^2)

2* series 1/(n^2) is a convergent p-series

p>1

1/(n!) is convergent by Comparison test

would this be right whatsoever, if not can somebody tell me the mistake im making. thanks a lot

[nilsoner1]

Thanks man! you are saving my ass in calc

[btran92]

this is one of the hardestest part of series to grasp imo haha

[birdsofafeather1990]

At 3:18, how is that divergent? A graph of y = 1/sqrt(x) seems to converge to 0.2

[travdoeshave88]

@birdsofafeather1990 remember with the test for divergence( lim x-->infinity f(x) = anything but 0 then it is divergent) if the limit converges that doesn't mean that the series converges.

if you continually ad .2 an infinite number of times then the sum would = infinity

.2+.2+.2+.2+.2=5(.2)

.2+.2+.2+.2.......+.2+.2=n(.2) for infinite series n goes to infinity so n(.2)= infinity

[sherlocklol]

doing this for ib paper 3... anyone else?

[omgaify]

I THINK I UNDERSTAND.

[kmin2]

You said sinx is between -1 and 1 so what is cosx in between?

[Tiboroun]

@kmin2 same range for cos.

[Nanaadjoa1990]

God Bless you, Patrick

[randomsubscriber100]

what grade is this taught

[mikegovikes]

cant you just use the ratio test? It would be way easier

[titan59]

@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

[MetalSio]

What about 1/n!  from zero to infinity

Thanks!

[RyuHanamichi]

I really enjoy your calm voice

[Jeiaga]

basically just had a nerdgasm when i finally understood how you choose an and bn. all because of you patrick, all because of you..

god i can't wait till calc is done.

[pippsful]

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

[spicyxlilxman]

it's so refreshing seeing a fellow left-hander teach math. this was a very helpful video. thanks.

[dajakesta1234]

but if u have the sumation wont it get closer to 2 at last???

[floyd617]

OKAY PATRICK I DONT GET IT!!

how did you cancel all the terms but leave the last two, the 2/n and the 1/n you said you cancelled the rest because they were smaller than one.. but so is 2/n and 1/n. Why did you leave those terms intact?

please answer quick test in 2 days :)

and thanks for all your help

[patrickJMT]

@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)

[AznVamp24]

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! :]

[patrickJMT]

@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.

[dimplezzali]

@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?

[dimplezzali]

@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!!!

[lastgoduni]

@AznVamp24 1/n is a harmonic series which is divergent...but 2/n^2 is convergent

[stealinglemons]

In which video do you cover "p-series"??

[NagashiChidorii]

@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 <= 1 ; Diverges

In the comparison test you compare against these basic series 1/n^p and if P fits any of the rules above you can conclude if it Converges or Diverges.

[aarontbh]

I just realized that I no longer do homework or study for tests without Patrick's help. Thank you Patrick!

[saltydog78]

If we prove the limit comparison test, can we refer to the sandwich theorem to prove it or is the sandwhich thereom totally different.

[Gemparkzz]

YOU ARE MY SAVIOUR........

[sassysafi]

Thank you so very much. My calculus teacher is out of control. Thank you so much.

[Cogwirrel]

What's going on in the denominator? 3:52

[Ayplus]

This video was beautifully constructed and such but unfortunately he didnt go over the situations where you attempt to compare the An with a Bn and the Bn you chose is larger than An but divergent or when the Bn you chose is smaller than An and converges. Thats usually when you get stuck and the problem gets superhard and your not sure where 2 go from there.

[Jawshooah]

Regarding the last example (n!/n^n), is there a more formulaic way to solve it? Your method is pretty intuitive, but it doesn't seem to adhere to any particular test until the end, where you use direct comparison with 2/n^2.

[lilb0ibastard]

Excellent vids. If I remember correctly, these were calculus 2 material, and though the series formula's were do-able, the main trouble comes from finding out which test/technique to use for the problem and why...

[loljkjklol]

Thank you for this :)

[asianconfusion]

your voice sounds different in this video at first i thought it was someone else

[patrickJMT]

@asianconfusion many years of making videos through many moods and levels of health.

[FaiththeHairstylist]

Ok, watched this again today. This was from the Stewart book, and I have the complete solutions manual, but STILL did not understand how to do it! Now it's starting to make more sense. Thanks again for your videos.

[FaiththeHairstylist]

haha, squeeze

[jisx90]

Can you please go to Umd and become my cal 2 professor please?

[tadm123]

1/n^3-4n^2

Why do some people compare this to n^2? if the highest power on bottom would be n^3? im confused

[452923]

what if you do a limit test and the result is 0. What is the the conclusion then??

[Tentenlaw]

@452923 : According to the limit comparison test, if the limit is greater than zero, then, both a (sub n) and b (sub n) both diverges or converges. If the limit is zero or lower, the test fails. You must use another test.

[452923]

@Tentenlaw Thanks man!

[fizzmaverick]

dude i love your videos man! thanks for the help your better than my prof and textbook combined!

[ConorD1990]

Thanks you very much. : )

[randumbtimes3]

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)

[tiofilo75]

Does anybody know why for the last problem of the n!/n^n, he write it out and the last one he leaves as 3*2*1? I am having trouble understanding that.

[greyfoxxelite]

patrickJMT is the god of calculus thank you so much for the help

[monoObabe]

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<0,0002

then we can say n!/n^n < 2/n * 1/n

I hope I'm right o.O

[cakeisreal]

i dont understand how n!/n^n< or equal to (2/n)(1/n).

[conclusion66]

thanks buddy

[luisaviation]

how do you know know when to use the limit comparison test or direct comparison test?

[waspy89]

It's so amazing that you do these. I'd be failing Calc 2 if it weren't for being able to review the material via these videos. Thanks!

[Payme4ril]

thanks for ur videos, when u say p series, do u mean power series? I'm a bit confused as I'm thinking of power series as things in the form of 1/(1-x)..am I wrong? thanks..

[troubadorra]

@Payme4ril power series =/= p-series, go to the Wiki about series in math and then go to Examples, then go to the one which mentions Riemann. instead of writing it as a form of 1/n^r just write 1/n^p and if p > 1 it converges.

[randumbtimes3]

@Payme4ril p-series

1/n^c

with c being a constant. if c > 1 it converges

if c <=1 it diverges

[Bry1226]

Awesome video

[pochankitty]

This guy is so great. Great teacher, and just seems like a damn decent guy; who else helps people this much for free? Respect, patrickJMT!

[Teenageriot280]

@pochankitty not free. dont you see the commercials? lol. great teacher nonetheless. definitely always rely on you, patrickJMT, when my teachers fail to teach

[kremesota]

great video.

[CominUpShort01]

why do you take the last two terms (2/n)(1/n) instead of just the last term.. or the last three terms? how do you know how many terms to take?

[Blzeheat09]

I have no idea either. Same question Patrick and thanks for the video!!

[mastermirza]

He takes the last two terms because he realizes that 2/n^2 is convergent since it is a p-series with p > 1, so that way he can compare the original series with 2/n^2 to find that it is convergent, by The Comparison Test

[Hawkallica]

Where did bn come from and how come he can just multiply it with the original problem?

[exteris]

@Hawkallica

Bn comes from An; Bn is a function that acts the same ways An acts. (You can also determine bn by taking the highest degree on the numerator and and diving it by the highest degree on the denominator)

 For example if An= 1/n+5 then your Bn=1/n but of course it wont be as simple as that.

The reason why he multiples Bn with An is because the limit comparison test states to do so.

lim n->infinity of An/Bn

[metalfreak66666]

nice explaining, i actually had this doubt lol.Weird how a comment helped me out.

[xxKrazy2nRxx]

Thanks for the vid, the part i'm stuck on is how do u know if the new series you make is supposed to be bigger or smaller than the original.

[Jakathera]

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.

[xSilverScorpio17x]

Couldn't you just put the constant inside parentheses and raise that quantity to the power of n, putting it into geometric form?

[Jakathera]

Good eye there xSilver.....

I have to agree with my math theacher... I'm not smart, just stubborn.

[maths486]

since ( 1 + sin n ) / 10^n is bounded, it means that it is convergent ? is that what you mean ?

[csusbkid]

alright. first problem, why is it for all n greater or equal to 1? does it have to be that? i would have thought that it would have been all n greater than or equal to 0.

[xpnoipj0230]

i don't mean to be rude, but why do you divide your a-sub-n with your b-sub-n in example 2?

[xpnoipj0230]

never mind, don't answer that. i just learned that limit comparisons, you are supposed to do that.

[therayne]

Woops, i replied, sorry I didn't see this until just now.

[therayne]

It's not being rude, it's ok to ask questions. =]

He divides a.n by b.n because that's the rule for Limit Comparison Test. Google Limit Comparison Test or consult your textbook and you'll see it.

[Nanumir]

thanks!

[peasncrackers]

i have a question...

if i wanted to, could i use the limit comparison test for all comparison problems? or is there a time to use the limit comparison versus the direct comparison test?

[gharib4ever]

I may be wrong...

If you can use the direct comparison test, there is no need for limit comparison,, check out Patrick's other vid, it explains it.. =D good luck

[TheLemon22]

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

[peglegjim]

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)?

[claratatatan]

ty patrickjmt

you just saved me 1 hour of reading through the difficult language of my calc book.

[jadenbane]

For what it's worth writing all the (n - k) / n as =< 1 is a little sloppy. Strictly speaking, the terms must be 0 < (n-k) /n < 1 in that product. The point is still made of course, but I feel as though that makes the comparison to 2 / n^2 more clear. Great video! Five stars.

[RadoAller]

In the last example why did you choose the last two numbers...(2/n and 1/n),,,,i still don't get it.>>>>

[TheV0iceOfReas0n]

hey man thanks so much, your videos are amazing

[Sarahbelle305]

I learn more from your ten minute videos than I do from a whole week of class! Thanks for the help!!

[musicloverboi]

I sure hope I have something like this for my exam! Then maybe only a handful of us will know of to do it! =D (Including me of course! =P)

[whatduwanna]

you r really great :) i got everything :)

come to be a lecturer in GAU :))

[shivamappadu]

At about 4.10min, i dont understand why you multiply by (n^(1/2))

[peasncrackers]

the formula for the limit comparison is lim as n-->infinity of (an/bn). so his bn was (1/(n^1/2)). dividing by 1/(n^(1/2)) is the same as multiplying by (n^(1/2)) (aka the reciprocal)

[seanzzy420]

I am taking calc II thanks for all the videos u make i would noy have made it this far without your help thanks pat

[pob11212]

would the ratio test also be useful for the last problem: n!/n^n ?

[panth92]

what happens if the limit equals zero

or if the limit is infinity????

[trawa35]

Dzieki stary pomogłes mi pass my exeam

[Senefleder55]

May I ask u something?

At the part around 3:10, u wrote root n/n = 1/n power half

I still don't understand how u got that

Thanks :)

[patrickJMT]

(n^(1/2))/n^1 = 1/n^(1/2) ; u are just subtracting exponents

[Senefleder55]

aha, got it!

Thanks! :)

[naumanahmed19]

@patrickJMT thanks

[naumanahmed19]

@patrickJMT thanks

[petersun825]

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.

[The4thbassguy]

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!

[03IMPALASS]

ok i see thanks man

[03IMPALASS]

whats the point of the limit comparison if you already found out that it was div by the p-series?? plz someone help

[patrickJMT]

the original series is not a p-series though

[03IMPALASS]

for your 2nd example you compared the original series to 1/n^1/2 which is div...so why the go the extra mile and do the limit comp?

[patrickJMT]

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'

[jg215]

u did the limit because it has to be > 0 and a finite number... correct?

[Chvoirre00]

@patrickJMT so is it ok if i use direct comparison for this question and not going the extra mile and do limit comparison?i still don't get what do you mean by 'hand waving'....

[The4thbassguy]

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).

[AllYourBase420]

Very helpful. Keep it up, I got a calc test coming up :P

[derek5218]

thank you so much, this is much more helpful than lecture

[javalu882005]

FXXX YEA. IAM going to pass my CACL Test, THANKS MAN!

[tguy1089]

i like the way you draw your sigmas

[Dynamics18]

was n=0 supposed to change to n=1?? lol ^_^

[toolrocks666]

factorials are a pain in the ass