Strategy for Testing Series - Series Practice Problems

[jeffschroeder43]

for number 4 I cant seem to get it to converge when I split it up I get 3((-1)^n)((3/8)^n) which goes to infinity when you test if the limit it goes to zero

[patrickJMT]

@jeffschroeder43 i think my intuition was wrong on one or two of them; you may read the other comments to see what others think

[ICarnag3I]

I think I'm gonna watch your calc 2 playlist before I start MV calc next semester = ]

[patrickJMT]

@ICarnag3I sounds like a not terrible idea ; )

[jamesa07]

Great video. Did you post solutions to these problem by any chance? :D If not, are you going to be making one?

[patrickJMT]

@jamesa07 no and probably not : ) i have a lot of sequence and series quizzes and solutions on nixty dot com if you are interested

[Billabong024]

i think number 8 diverges. I got the limit to equal 2.

[porkystarcraft]

@Billabong024 I got the same answer as you did.

[plbernstein92]

@porkystarcraft if i'm not mistaken, since it diverges by the ratio test it could still conditionally converge. Patrick, can you please address this one: my test is tomorrow morning!

[porkystarcraft]

@plbernstein92 It depends if the original series is alternating or not i think. Since the ratio test already takes the abs value of the series it solves for all the positive values...if the series before abs value is taken is still all positive there is no chance of it being conditionally convergent....my calc 2 final is wed.

[Billabong024]

DUDE I don't even need to do HW from my textbook anymore LOL.

[lllXchrisXlll]

I am getting the #2 to diverge. I tried pulling out the "n" and splitting the fraction. This gave me to series where one was a p-series with n^1 wich diverges. I also tried using the ratio test, but that also diverged. Anyone have any tip they could share?

[BMChicago]

@lllXchrisXlll When you pull out an n, then you should be left with 1/n^2. Basically you get rid of the + n. Then you use the 1/n^2 to compare with the original series. 1/n^2 is a convergent P series because p=2 > 1. Then use the limit comparison test.

[lllXchrisXlll]

@BMChicago Thanks, i messed up my algebra trying to do the ratiotest. The direct comparison test is defenatly much better. Thanks.

[schappell08]

This was such a good video for review! thank youuuu!!!!



[Reedoc3]

You inspire me to stick with my major in Mathematics.

[patrickJMT]

@Reedoc3 wonderful : ) 

[shariffhazim11]

hi! great videos!

I dont get #11! and 12. You have som video of this?

[jonnysmpr]

6:51 "make this SHIT disappear" LOL

[guitardude006]

Patrick, I fucking love you. My Calc II prof is pretty much sitting through her tenure, not at all interested in helping us learn the material. I wish there were more math teachers like you. Thanks for the test prep!

[KingLiamYNWA]

very good video. . i was asked to find the sum of a series in a question, and its nt geometric, anyone know how to go about that?

[Edisonstar23]

When worked out #8 w/ the ration test, I ended up w/ =2, So since 2>1 it diverges!

Please someone confirm or correct me 

[nubxy]

@Edisonstar23

I think that is correct. I also got 2. And therefore it diverges.

[SirDontTouchMeThere]

I think you're entitled to a portion of my professor's salary....

[styxfreak]

thanks so much for this, i wish more people did things exactly like this. its a really great way to see if you actually understand where to start- which is normally the problem

[LFZ15]

so great, I need to learn how to get a good grasp on series pretty quickly and just don't have the time to work the dozens of problems to get a good intuition of them like you're supposed to, this video was a lifesaver. thank you SO much.

[Q8iEviL]

Thumbs up if u think patrick ROCKS!!!

[The12thAkatsuki]

For question number 8, you can write that as 2^k divided by (k+2)(k+1) right? I think the problem becomes easier that way. And I think it's divergent because the limit is 2. Correct me if I'm wrong though. :)

[Wilhemet]

For #8, when I did the ratio test, I ended up with 2K! , which makes it diverge?

[patrickJMT]

@Wilhemet it is possible, i did not check them

[55movie55]

@Wilhemet i got 2k/k which s 2 which is >1 so divergent

[gebzeliomer]

yeah me too

[03kingh]

Patrick, I've been a fan of your videos for a few years now and have to give you props for the great work. Now, I'm soo lost. Im using Stroud at the moment and i can't access the solution for this question. I'm stuck. ANY HELP AT ALL would be much appreciated. The question is prove that 1+2x/5+(3x^2)/25 +(4x^3)/125+... is convergent for -5<x<5 and for no other values of x. In need of help. Thanks

[I3urningi3laze]

By the ratio test, #8 should diverge because the resulting limit is 2 which is greater than 1.

[ny1fanta]

for #13, I keep getting infinity/e after i do it with ratio test. so the limit is divergent?

[Tofz]

@ny1fanta yeah i keep getting the same thing

[archiemedes42]

Are you sure about 8? I think, since the factorials just simplify to get 2^k / (k+2)(k+1) with an exponential numerator and a quadratic denominator, that has to diverge, right?

[zvcxk]

@archiemedes42 I have the same question... 

[godspeed28]

fan fucking tastic

[patrickJMT]

@godspeed28 ha

[marichich]

I think that number 11 is incorrect. It is diverge. |sine 1/n| / n^(1/2) diverges by direct comparison test with diverges p-series , where p =1/2<1. |sine1/n| / n^(1/2) <= 1/ n^(1/2) since |sine1/n| <= 1

[patrickJMT]

@marichich that is incorrect. showing that a series is smaller than a known divergent series does not prove that it the original also diverges.

[stoyashot]

@patrickJMT But what if you would compare number 11 with sin(1/n)/n a divergent series, number 11 will be bigger than sin(1/n)/n because n^.5 > n . With this comparison number 11 diverges. So: 0 < sin(1/n)/n < sin(1/n)/(n^.5) ; sin(1/n)/n diverges: p-series and so does number 11 by comparison.

Tell me what you think.

[scottandrus92]

@patrickJMT So in that case, you would use the limit comparison test?

[FaiththeHairstylist]

Great video. I'm having such a hard time with series. I have the complete solutions to Stewart, but it doesn't always help. Hearing your actual thought process is what I needed.

[Dmaqur91]

for #7, can you do a comparison test and pull out 1/n so that you get a p-series that equals 1 which makes it divergent? Not really sure about that one

[patrickJMT]

@Dmaqur91 i am not quite sure what you mean; the direct comparison test and limit comparison test will be inconclusive if you compare to 1/n, i do believe.

[mishimmy1]

Any clues on how to tackle the series (cosn)^2 / (n^2 +1) by using a suitable test to determine convergence/divergence would be much appreciated :)

[patrickJMT]

@mishimmy1 show it is absolutely convergent and bound the new series by 1/(n^2 + 1) which implies convergence by the comparison test and p-series.

[mishimmy1]

@patrickJMT thank you so so so so much!!!!

[charlesc300]

i think 12 can also be evaluated by using root test they all have the same exponential k, i just think is way easier.

[alexcuenca]

Thank you very much! keep up the good the good work!

[meltedcup]

You're the bomb! These videos make math much clearer. Unlike some people, I love how you explain each step as to ensure our success in the topic :) And it's clearly working! I'm rocking my assignments. Thank you thank you thank you !!!

[DaGuyFlyinHigh]

Less stuff about describing what you're doing (Ex. So this is just to show you about the series, and stuff to do, and we're going to talk about this, and that, and then once again this is about series.....) get to the math, and don't explain it like you're teaching retards. You don't have to teach us what a limit is over again, if we're trying to know what the limit actually values. But other than that, its good stuff.

[patrickJMT]

@DaGuyFlyinHigh i think i will probably keep doing exactly what i want to do.

[itheds]

@patrickJMT it is the freedom of your methods that make your videos so valuable. Keep up the good work, and thank you.

[yusufforever]

@patrickJMT hell yeah bro, you have helped and continue to help me with mathematics and the way you explain is by far the best, next to an actual face to face professor. i still have at least 2 and a half years until i get my degree so please don't change.

[george21f]

@DaGuyFlyinHigh beggers can't be choosers.

[jayorca]

Your videos are very helpful.

[patrickJMT]

@jayorca glad you like 'em.

[gsowx2]

Found it! Awesome, thanks

[karineattar123]

you are amazing

[karineattar123]

very helpful thanks a lot

:)

[uberuberdude101]

i fucking love you....this video helped a ton

[TheElephantShoe]

thanks for all your videos, you are the only reason why i survived calc ii. =]

[torment0fsin]

increase the volume for your videos

[13loodhound17]

@torment0fsin plz

[patrickJMT]

@torment0fsin ok : )

[mohemedamin]

2 > 1 so it diverge not converge according to the Ratio test

[ShipLoot]

This was soooooo helpful, i can do just about every problem from the book when you are only using a certain method but I've been struggling when it comes to picking the correct method to use for random problems on quiz's/exams. This video was a big help for me.

[Haengma]

I think you're right, it looks like the ratio test says it should converge if the limit goes to a value greater than 1.

[atlvietnguyen]

number 11 should be diverging. u can't use a direct comparison with that because 1/n^3/2 is less than (sin(1/n))/(n^1/2).

[Axus1]

Should diverge if greater than 1, is what I think you meant to say.

[spydr11]

You are correct.

[KittyBun1]

Your videos helped me out sooooo much. Thanks and keep them coming. I learn sooooo much from them. :)

[woodengun]

hey bro, thanks a lot.

[BadDrafter]

Thanks for the A on the mid term! I memorized the shapes of the series' and just was automatic on the test.

[jayteacher]

Wish I had seen this video for limit tests before I took my college Calc 2 course final exam. My grade would have definitely been better. The professor did not explain to his students how to start to examine the limit problems to find the best test(s) to use. He already knew that test to use, so he would tell us that we would use this test or a combination of tests to prove convergence, divergence, or unknown. If unknown, he would begin another test, but most students did not get the strategies.

[mary89xmc588]

hehe.... converge or diverge compared to what ? =p silly question but i do not know.

[ale06053]

thanks a lot mate ! you help me a lot..

[HamsAlamal]

wow great work teacher

Now I can know how to start testing series

thank you alot

[TAHMURI]

Your the best!

[atomo360]

I guess i'm gonna wait until I see aleast something that looks like this... :D

[damianwba]

:-)