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

1. anonymous

$\sum_{1}^{\infty} \frac{ n ^{2} }{ 2^{n} }$

2. anonymous

@ayeshaafzal221

3. IrishBoy123

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

4. IrishBoy123

Does $\lim\limits_{n→∞} a_n = 0$

5. anonymous

I'm trying to reach the final result

6. anonymous

I wanna see how to do that

7. 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:

8. IrishBoy123

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

9. 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

10. anonymous

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

11. anonymous

derivate*

12. IrishBoy123

cool do it!

13. anonymous

But (n^2)' = 2n

14. IrishBoy123

yes do the bottom next

15. anonymous

and I dont know how to derivate 2^n

16. Jhannybean

$\sf a^x = a^x \cdot \ln a$

17. Jhannybean

oop... $$\frac{d}{dx}$$* of..... haha.

18. anonymous

Oh

19. anonymous

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

20. anonymous

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

21. IrishBoy123

yes

22. anonymous

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

23. anonymous

24. IrishBoy123

so what does that prove?

25. anonymous

It is convergent?

26. IrishBoy123

no!!!

27. anonymous

Dammit 50:50 chance and I failed

28. 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

29. Jhannybean

p-series?

30. IrishBoy123

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

31. anonymous

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

32. Jhannybean

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

33. Jhannybean

Ahh, i meant not greater than.

34. Jhannybean

There is no darn symbol for that.

35. 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

36. IrishBoy123

@xfire30 you there and OK ?!

37. 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?

38. IrishBoy123

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

39. IrishBoy123

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

40. IrishBoy123

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

41. 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

42. 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

43. anonymous

No,it is bigger than 1

44. IrishBoy123

you sure? $\large \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}$

45. anonymous

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

46. IrishBoy123

this is the ratio test you were going to apply

47. anonymous

I dont get it

48. anonymous

How did you got that?

49. IrishBoy123

|dw:1442256369176:dw|

50. IrishBoy123

|dw:1442256441060:dw|

51. IrishBoy123

work that through to a conclusion :p