Solve this math problem by midnight PST and you get all my protocoin

find if this series converges and diverges. show your steps and what test you used.

I know it converges. just tell me the steps bro

This problem is actually pretty easy; once you get the hang of how to do them it'll be very easy for you to do. In a convergent infinite summation series, we know that the expression getting summed up must converge if the limit of the expression (1*(1/(n^sq(n)-n))) has a limit of zero. In the long run, this would mean that the expressions, for each additional value of n, would continuously become smaller and smaller, and thus the summation would eventually end up being a finite number. We know that n^sq(n)-n will keep increasing, as n^sq(n) as a higher exponent than n (>1 as compared to 1), and thus in the long run, n^sq(n) will be infinitely larger than n. The expression 1/(n^sq(n)-n) will thus eventually reach a limit of 0 in the long run, and because multiplying that value by 1 returns the same value, you know that the complete expression (1*(1/(n^sq(n)-n))) will have a limit of 0 in the long run. As per the Limit Comparison Rule, we now know that the series in the problem must converge.

Alternatively, we could have used the ratio test in order to compare (a_n+1/a_n) of the expression. Because a_n+1 must always be lower than a_n (note that over time the expression gets constantly smaller and smaller), we know that the ratio of those is under 1, which is only true if a series is convergent. Thus, the series in the problems must be convergent.

Here's a pdf link with convergence/divergence rules if you need: math.harvard.edu/archive/1b_spring_04/exams/exam2_review.pdf

Btw are you in AP Pre-Calc or Calc? Your problem looks like something that was taught in 11th grade and maybe 12th grade

Answer is 42.

נכתב על ידי clovis1122, 22.04.2018 at 20:36

Answer is 42.

Respect

נכתב על ידי Player 999, 22.04.2018 at 20:08

This problem is actually pretty easy, once you get the hang of how to do them it'l be very easy for you to do. In a convergent infinite summation series, we know that the expression getting summed up must converge if the limit of the expression (1*(1/(n^sq(n)-n))) has a limit of zero. In the long run, this would mean that the expressions, for each additional value of n, would continuously become smaller and smaller, and thus the summation would eventually end up being a finite number. We know that n^sq(n)-n will keep increasing, as n^sq(n) as a higher exponent than n (1.5 as compared to 1), and thus in the long run, n^sq(n) will be infinitely larger than n. The expression 1/(n^sq(n)-n) will thus eventually reach a limit of 0 in the long run, and because multiplying that value by 1 returns the same value, you know that the complete expression (1*(1/(n^sq(n)-n))) will have a limit of 0 in the long run. As per the Limit Comparison Rule, we now know that the series in the problem must converge.

Alternatively, we could have used the ratio test in order to compare (a_n+1/a_n) of the expression. Because a_n+1 must always be lower than a_n (note that over time the expression gets constantly smaller and smaller), we know that the ratio of those is under 1, which is only true if a series is convergent. Thus, the series in the problems must be convergent.

Here's a pdf link with convergence/divergence rules if you need: math.harvard.edu/archive/1b_spring_04/exams/exam2_review.pdf

Btw are you in AP Pre-Calc or Calc? Your problem looks like something that was taught in 11th grade and maybe 12th grade

good shit. you have Bn = 1/n^sqrt(n) and use the LCT. i missed the fact that n^sqrtn rises faster than n and kept using n as the Bn. ok, ask mod/admin to transfer all of my proto to your account and i'll approve. this was a question from a quiz for calc 2 from college.

mods u can lock this thread idc

נכתב על ידי clovis1122, 22.04.2018 at 20:36

Transfer the protocoin plz

But the question is, when n=1, 1/(n^sqrt(n) - n)=1/0=???

נכתב על ידי EtaoinWu, 22.04.2018 at 22:03

But the question is, when n=1, 1/(n^sqrt(n) - n)=1/0=???

Infinity; but the point is that the limit eventually reaches 0 for that part of the expression. 1/(n^sqrt(n) - n) does equal infinity (honestly the math professor should have done n = 2 lol) but it just matters that the expression gets smaller and smaller for increasing values of n

נכתב על ידי Player 999, 22.04.2018 at 22:11

Infinity; but the point is that the limit eventually reaches 0 for that part of the expression. 1/(n^sqrt(n) - n) does equal infinity (honestly the math professor should have done n = 2 lol) but it just matters that the expression gets smaller and smaller for increasing values of n

But any finite sum of the series is infinity right? (didn't read any math works in English, I'm not sure if the wording is correct)

The trick here is that you have 0 / very few protocoins.

נכתב על ידי EtaoinWu, 22.04.2018 at 22:18

נכתב על ידי Player 999, 22.04.2018 at 22:11

Infinity; but the point is that the limit eventually reaches 0 for that part of the expression. 1/(n^sqrt(n) - n) does equal infinity (honestly the math professor should have done n = 2 lol) but it just matters that the expression gets smaller and smaller for increasing values of n

But any finite sum of the series is infinity right? (didn't read any math works in English, I'm not sure if the wording is correct)

no that is not true

31

420

נכתב על ידי clovis1122, 22.04.2018 at 20:36

Answer is 42.

dont mind me give me a free info but isn't the value getting closer and closer to zero as N goes from n=1 to n to infinity? for my logic the answer is 0 and the number output will by no way be more then one* because the dominator is increasing so how do you get 42

נכתב על ידי SyrianDevil, 23.04.2018 at 10:17

נכתב על ידי clovis1122, 22.04.2018 at 20:36

Answer is 42.

dont mind me give me a free info but isn't the value getting closer and closer to zero as N goes from n=1 to n to infinity? for my logic the answer is 0 and the number output will by no way be more then zero because the dominator is increasing so how do you get 42

נכתב על ידי clovis1122, 22.04.2018 at 20:36

Answer is 42.

How did my real answer only get 1 upvote while a fake answer got 18 :O

Ive forgotten how to do those. But hint: google and youtube are your friend.

>When you see your own name as last comment on a promoted thread, and think to yourself ''who the fuck hacked my account''
> Opens the thread
> Oooh...

D

That = Chinese food

נכתב על ידי Permamuted, 23.04.2018 at 10:29

Ive forgotten how to do those. But hint: google and youtube are your friend.

google and youtube can't solve it.

נכתב על ידי boywind2, 23.04.2018 at 19:31

נכתב על ידי Permamuted, 23.04.2018 at 10:29

Ive forgotten how to do those. But hint: google and youtube are your friend.

google and youtube can't solve it.

This is a pretty standard problem. You can find both the theory and worked problems similar to this on youtube u lazyarse.

נכתב על ידי boywind2, 23.04.2018 at 19:31

נכתב על ידי Permamuted, 23.04.2018 at 10:29

Ive forgotten how to do those. But hint: google and youtube are your friend.

google and youtube can't solve it.

www.khanacademy.com is too op for math

I'm a 0 at math, but any History/Juridical stuff hit me up

נכתב על ידי Permamuted, 23.04.2018 at 22:11

נכתב על ידי boywind2, 23.04.2018 at 19:31

נכתב על ידי Permamuted, 23.04.2018 at 10:29

Ive forgotten how to do those. But hint: google and youtube are your friend.

google and youtube can't solve it.

This is a pretty standard problem. You can find both the theory and worked problems similar to this on youtube u lazyarse.

no u cant ive tried to find for an hour. the biggest point i missed was that for some reason i used 1/n as bn for the lct. i just reread the question and i realized thatvthe question was 1/nsqrtn - n not 1/n^sqrtn - n which doesnt make any difference since 1/n^sqrtn is still convergent but i just spent 2 hours on a question cuz i read it wrong twice fuck me and cexpert or whatever ur name is you didnt read my question right either because i noticed u said that n^sqrtn is 1.5 larger than n which is not true nsqrtn is but in the end it doesnt really natter since they are literally the same shit hopefully my prof doest realize that i actually wrote the question as n^sqrtn instead of nsqrtn since they both converge by the lct and the convergence of bot^ bn is proven using the ratio test fuck im rambling fuck you