Question

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

Answer 1

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

Answer 2

@ayeshaafzal221

Answer 3

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

Answer 4

Does \[\lim\limits_{n→∞} a_n = 0\]

Answer 5

I'm trying to reach the final result

Answer 6

I wanna see how to do that

Answer 7

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:

Answer 8

sorry it looks good as polynomial on top, exponent on bottom, my bad

do you know l'Hospital?

Answer 9

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

Answer 10

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

Answer 11

derivate*

Answer 12

cool

do it!

Answer 13

But (n^2)' = 2n

Answer 14

yes

do the bottom next

Answer 15

and I dont know how to derivate 2^n

Answer 16

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

Answer 17

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

Answer 18

Oh

Answer 19

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

Answer 20

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

Answer 21

yes

Answer 22

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

Answer 23

this is the answer?

Answer 24

so what does that prove?

Answer 25

It is convergent?

Answer 26

no!!!

Answer 27

Dammit 50:50 chance and I failed

Answer 28

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

Answer 29

p-series?

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Ahh, i meant not greater than.

Answer 34

There is no darn symbol for that.

Answer 35

@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

Answer 36

@xfire30

you there and OK ?!

Answer 37

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?

Answer 38

if \[ \lim\limits_{n→∞} a_n \ne 0 \]

it diverges

Answer 39

but if
\[\lim\limits_{n→∞} a_n = 0\]

on we plod, it could still go either way

Answer 40

so we need another test

can you try this one from the sheet?

RATIO TEST
Is limn→∞ |an+1/an| 6= 1?

Answer 41

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

Answer 42

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

Answer 43

No,it is bigger than 1

Answer 44

you sure?

\[\large \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\]

Answer 45

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

Answer 46

this is the ratio test you were going to apply

Answer 47

I dont get it

Answer 48

How did you got that?

Answer 49

work that through to a conclusion :p