I have a issure with a Series exercises,anyone can help please?

- anonymous

I have a issure with a Series exercises,anyone can help please?

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- anonymous

\[\sum_{1}^{\infty} \frac{ n ^{2} }{ 2^{n} }\]

- anonymous

@ayeshaafzal221

- IrishBoy123

are you trying to establish whether the series converges/diverges?

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## More answers

- IrishBoy123

Does \[\lim\limits_{n→∞} a_n = 0\]

- anonymous

I'm trying to reach the final result

- anonymous

I wanna see how to do that

- 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

sorry it looks good as polynomial on top, exponent on bottom, my bad
do you know l'Hospital?

- 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

Yeah,the basics,I derivated each term and try to do the limit again

- anonymous

derivate*

- IrishBoy123

cool
do it!

- anonymous

But (n^2)' = 2n

- IrishBoy123

yes
do the bottom next

- anonymous

and I dont know how to derivate 2^n

- Jhannybean

\[\sf a^x = a^x \cdot \ln a\]

- Jhannybean

oop... \(\frac{d}{dx}\)* of..... haha.

- anonymous

Oh

- anonymous

so it's (2n)/(2^n) * ln 2?

- anonymous

and it is infinite / infinite and I do lhospital again?

- IrishBoy123

yes

- anonymous

0/(2^n)*ln 2 = 0

- anonymous

this is the answer?

- IrishBoy123

so what does that prove?

- anonymous

It is convergent?

- IrishBoy123

no!!!

- anonymous

Dammit 50:50 chance and I failed

- 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

p-series?

- IrishBoy123

An example: \[\Sigma \frac{1}{n} \] diverges even though the individual terms get smaller and smaller

- anonymous

I see,because the "p" equals 1,right?

- Jhannybean

Oh because according to the p-series test, \(\dfrac{1}{n^{\color{red}{1}}} \) means that 1 \(\gt\) 1, so it diverges.

- Jhannybean

Ahh, i meant not greater than.

- Jhannybean

There is no darn symbol for that.

- 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

@xfire30
you there and OK ?!

- 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

if \[ \lim\limits_{n→∞} a_n \ne 0 \]
it diverges

- IrishBoy123

but if
\[\lim\limits_{n→∞} a_n = 0\]
on we plod, it could still go either way

- IrishBoy123

so we need another test
can you try this one from the sheet?
RATIO TEST
Is limn→∞ |an+1/an| 6= 1?

- 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

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

No,it is bigger than 1

- 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

I dont get it,is this another exercise or the same one?

- IrishBoy123

this is the ratio test you were going to apply

- anonymous

I dont get it

- anonymous

How did you got that?

- IrishBoy123

|dw:1442256369176:dw|

- IrishBoy123

|dw:1442256441060:dw|

- IrishBoy123

work that through to a conclusion :p

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