The Limit Comparison Test - Proof

Course Web Page: sites.google.c...

[@drhubblebubble7]

Damn, very well worded and concise. Thanks a lot for making this proof clear, I feel as if the calculus teachers I've had really speed through series too much and don't put in the time to clearly and simply show where these tools come from.

[@lukepaluso9863]

I agree, I've just drawn the conclusion that it is up to me to satisfy my own curiosities, it's unfair (in my opinion) to expect any program designed to teach numerous people at once to satisfy my curiosities of a topic. There is always room for me to want to learn more

[@hizircanbayram9898]

I've just wanted to thank you to meet us these stimulating videos!

[@jorgeluisrodriguez3271]

So if I understood the video at 4:39, to make boundaries for the value of the any series, we must prove term by term that: a_10 < L · b_10 ; a_11 < L · b_11 ; a_12 < L · b_12 ; ...; a_n < L · b_n; Sum a_n

[@AryaDhayal]

thankyou :)

[@microhoarray]

lot of teachers just gives you the textbook proof and read it out loud while thinking that they're explaining the stuff. reading out loud is not an explanation, but this is! that you for showing it it visually and not mentioning the damn epsilon-delta definition.

[@slcmathpc]

I agree with your sentiment. Although absolute rigor is eventually important, it should definitely come second after developing a solid intuition, which is far more important. Keep up the quest for true understanding! :-)

[@riley4051]

Thanks for this proof and I also wanted to note: I think you could evade the for n>N because you could set the interval between two points that (A subn)/(bsubn) never gets to which could be some negative value (A) and some positive real number (B) which would still yield accuracy to the proof.

[@slcmathpc]

No. It is absolutely necessary that the interval around the limit value k>0 containing a_n/b_n for n > N has positive endpoints otherwise the direct comparison does not apply when Sum b_n diverges and so the proof breaks down. This can only be guaranteed by taking N sufficiently large.

[@superoxidedismutase5757]

perfect explanation

[@riley4051]

question: why can't you sum from n=1 to infinity instead of n = (big N) to infinity? I understand that we have confined Asub n/B sub n to whatever interval a to b, but I don't think it would affect the statement's validity to sum from n=1 to infinity of both sides.

[@Rahul-bq6xl]

Thanx It cleared my concept very well Keep up the good work

[@lorenzoortiz8966]

thank you so much!!! you are a great teacher, i wish more PAID teachers had the will to teach that youtubers have. it's a shame because these bastards get paid to teach. i've had nothing but teachers that don't really like to teach mathematiccs. they're like, OK MAGGITS,, memorize this formula, plug , chug, and move on. mathematics is fun and interesting if you understand the inner workings.. otherwise it's nothing but self torture

[@kjlive781]

Thank you

[@BL-om2hn]

ur amazing love you

[@Imazine1104]

Well explained!!

[@keithlyons8931]

Well done!

[@michelazar1997]

u r amezing

[@aakarshikasingh4715]

You have mentioned that sigma an is bounded and so convergent. But this is not always true. A sequence needs to be monotone as well as bounded to be convergent.

[@slcmathpc]

Since a_n >0, then the sequence of partial sums is increasing.

[@nsq2487]

Can an and bn be negative? Will this test still work?

[@slcmathpc]

a_n and b_n must be eventually positive, since the test relies on the Direct Comparison Test.

[@santosshresth4419]

thank u sir greatly helpful

[@naif277]

My issue with this proof is that you assumed that the value ot the limit can get bigger than k which is impossible! Because k is the limit when n approach infinity. I know this video is old but can you please clarify this to me

[@slcmathpc]

You are confusing the individual terms of the sequence with the limit of the sequence. Consider the following two examples: 1+1/n > 1 for any n>0 and 1+1/n approaches 1 as n tends to infinity. 1-1/n < 1 for any n>0 and 1-1/n approaches 1 as n tends to infinity.

[@nasi5813]

I still dont get it fml

[@slcmathpc]

Keep at it, we've all been there!

[@michelazar1997]

i lav u

[@michelazar1997]

+Michel Azar i lav u too

[@michelazar1997]

+Michel Azar me too

[@sashalo1194]

@@michelazar1997 Wow this was a very odd but wholesome set of three comments. Welcome to the internet friends.