anonymous
  • anonymous
I have a issure with a Series exercises,anyone can help please?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sum_{1}^{\infty} \frac{ n ^{2} }{ 2^{n} }\]
anonymous
  • anonymous
@ayeshaafzal221
IrishBoy123
  • IrishBoy123
are you trying to establish whether the series converges/diverges?

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IrishBoy123
  • IrishBoy123
Does \[\lim\limits_{n→∞} a_n = 0\]
anonymous
  • anonymous
I'm trying to reach the final result
anonymous
  • anonymous
I wanna see how to do that
IrishBoy123
  • IrishBoy123
you 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
  • IrishBoy123
sorry it looks good as polynomial on top, exponent on bottom, my bad do you know l'Hospital?
IrishBoy123
  • IrishBoy123
i 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
  • anonymous
Yeah,the basics,I derivated each term and try to do the limit again
anonymous
  • anonymous
derivate*
IrishBoy123
  • IrishBoy123
cool do it!
anonymous
  • anonymous
But (n^2)' = 2n
IrishBoy123
  • IrishBoy123
yes do the bottom next
anonymous
  • anonymous
and I dont know how to derivate 2^n
Jhannybean
  • Jhannybean
\[\sf a^x = a^x \cdot \ln a\]
Jhannybean
  • Jhannybean
oop... \(\frac{d}{dx}\)* of..... haha.
anonymous
  • anonymous
Oh
anonymous
  • anonymous
so it's (2n)/(2^n) * ln 2?
anonymous
  • anonymous
and it is infinite / infinite and I do lhospital again?
IrishBoy123
  • IrishBoy123
yes
anonymous
  • anonymous
0/(2^n)*ln 2 = 0
anonymous
  • anonymous
this is the answer?
IrishBoy123
  • IrishBoy123
so what does that prove?
anonymous
  • anonymous
It is convergent?
IrishBoy123
  • IrishBoy123
no!!!
anonymous
  • anonymous
Dammit 50:50 chance and I failed
IrishBoy123
  • IrishBoy123
we 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
Jhannybean
  • Jhannybean
p-series?
IrishBoy123
  • IrishBoy123
An example: \[\Sigma \frac{1}{n} \] diverges even though the individual terms get smaller and smaller
anonymous
  • anonymous
I see,because the "p" equals 1,right?
Jhannybean
  • Jhannybean
Oh because according to the p-series test, \(\dfrac{1}{n^{\color{red}{1}}} \) means that 1 \(\gt\) 1, so it diverges.
Jhannybean
  • Jhannybean
Ahh, i meant not greater than.
Jhannybean
  • Jhannybean
There is no darn symbol for that.
IrishBoy123
  • IrishBoy123
@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
  • IrishBoy123
@xfire30 you there and OK ?!
anonymous
  • anonymous
Yes,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
  • IrishBoy123
if \[ \lim\limits_{n→∞} a_n \ne 0 \] it diverges
IrishBoy123
  • IrishBoy123
but if \[\lim\limits_{n→∞} a_n = 0\] on we plod, it could still go either way
IrishBoy123
  • IrishBoy123
so we need another test can you try this one from the sheet? RATIO TEST Is limn→∞ |an+1/an| 6= 1?
anonymous
  • anonymous
I 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
  • IrishBoy123
is \[ \lim\limits_{n→∞} |\frac{a_{n+1}}{a_n}| \lt 1\]??? i have to go too but try this out and see where you go
anonymous
  • anonymous
No,it is bigger than 1
IrishBoy123
  • IrishBoy123
you 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
  • anonymous
I dont get it,is this another exercise or the same one?
IrishBoy123
  • IrishBoy123
this is the ratio test you were going to apply
anonymous
  • anonymous
I dont get it
anonymous
  • anonymous
How did you got that?
IrishBoy123
  • IrishBoy123
|dw:1442256369176:dw|
IrishBoy123
  • IrishBoy123
|dw:1442256441060:dw|
IrishBoy123
  • IrishBoy123
work that through to a conclusion :p

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