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anonymous
 one year ago
I have a issure with a Series exercises,anyone can help please?
anonymous
 one year ago
I have a issure with a Series exercises,anyone can help please?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \frac{ n ^{2} }{ 2^{n} }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1are you trying to establish whether the series converges/diverges?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1Does \[\lim\limits_{n→∞} a_n = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to reach the final result

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wanna see how to do that

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you are adding the terms in a series but if they are increasing as you go there is no way the series can converge to a finite sum so look at the individual term. it just looks bad but you can apply l'Hopital's rule:

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sorry it looks good as polynomial on top, exponent on bottom, my bad do you know l'Hospital?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i am looking at this list, as you seem to need some technique or approach i'd suggest we just work down it http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah,the basics,I derivated each term and try to do the limit again

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes do the bottom next

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I dont know how to derivate 2^n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sf a^x = a^x \cdot \ln a\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oop... \(\frac{d}{dx}\)* of..... haha.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it's (2n)/(2^n) * ln 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and it is infinite / infinite and I do lhospital again?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so what does that prove?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dammit 50:50 chance and I failed

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1we applied the test to the individual term of the series and so the terms get smaller as this series goes on but it does not follow that the series is finite look at the link again http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf we have applied the first test. it can tell you something is not convergent but it cannot tell you whether something is convergent

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1An example: \[\Sigma \frac{1}{n} \] diverges even though the individual terms get smaller and smaller

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see,because the "p" equals 1,right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh because according to the pseries test, \(\dfrac{1}{n^{\color{red}{1}}} \) means that 1 \(\gt\) 1, so it diverges.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, i meant not greater than.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is no darn symbol for that.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@Jhannybean i'm literally blagging it from that sheet i linked, and it says Does \(a_n = 1/n^p, n ≥ 1?\), but we have Does \(a_n = n^2/n^p, n ≥ 1?\) i'd ratio test it as i feel safer

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@xfire30 you there and OK ?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes,just to make sure,if the result was 0,it was divergent no matter the type of the series,but if it is 0,it depends on the type of the series,right?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if \[ \lim\limits_{n→∞} a_n \ne 0 \] it diverges

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1but if \[\lim\limits_{n→∞} a_n = 0\] on we plod, it could still go either way

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so we need another test can you try this one from the sheet? RATIO TEST Is limn→∞ an+1/an 6= 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will try,But I have to go for 20 minutes to pick up my broather from the train station,I will tag you when I'm back

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1is \[ \lim\limits_{n→∞} \frac{a_{n+1}}{a_n} \lt 1\]??? i have to go too but try this out and see where you go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No,it is bigger than 1

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you sure? \[\large \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont get it,is this another exercise or the same one?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1this is the ratio test you were going to apply

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you got that?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442256369176:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442256441060:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1work that through to a conclusion :p
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