## tiffanymak1996 3 years ago is this absolutely convergent or conditionally or divergent?

1. tiffanymak1996

$\sum_{n=1}^{\infty} (-1)^{n}\times e ^{\frac{ 1 }{ n }}$

2. Herp_Derp

Divergent...

3. tiffanymak1996

yes that's the answer given, but i got conditionally throughout the process can u go thru the steps with me?

4. Herp_Derp

Well, for some large $$n$$, $$e^{1/n}$$ is sufficiently close to $$1$$ so that we can compare this series to the divergent series:$\sum_{n=0}^{\infty}(-1)^n$The larger the $$n$$, the closer your series is to this series. Therefore your series diverges. Do you understand why $$\sum_{n=0}^{\infty}(-1)^n$$ diverges?

5. tiffanymak1996

but, i thought n=1 under the sigma sign...

6. godorovg

No, tiffany they are telling right here, it is diverges

7. jeanlouie

we have abs($(-1)^{n}\times e ^{1/n}$) -> 1 when $n \rightarrow \infty$ but if consider $a_{n}$ = $(-1)^{n}\times e ^{1/n}$ , then |$a_{2n}$ - $a_{2n+1}$| > 1 with every n > some N it is diverges

8. tiffanymak1996

yeah, what i meant was i know it is divergent, but when i go thru the steps i got conditionally convergence

9. tiffanymak1996

don't understand this part an = (−1)n×e1/n , then | a2n - a2n+1 | > 1 with every n > some N it is diverges

10. jeanlouie

i mean, for example, let N = 1000000, then for every n > N, |a_2n - a_(2n+1)| > 1

11. godorovg

Jeanlouie I haven't this type calc for a while since High School, in order for this to be diverges it means the n is closer to 1 or I am lost here?

12. tiffanymak1996

for the ratio test if the lim is <1 then the series converges, if lim >1 or=infinity the series diverges, if lim=1 inconclusive

13. tiffanymak1996

i know that testing the series for absolutely convergent or conditionally or divergent is to try the ratio test with absolute value first then the one without.

14. tiffanymak1996

but when i calculate the one with absolute value the limit =1 so inconclusive; when i test the series for the ratio test without absolute value the limit =-1, so it's convergent; that's how i get conditionally convergent.

15. tiffanymak1996

although i know i'm wrong, cause the answer is divergent:(

16. godorovg

what is the value {1} of this when testing ratio =

17. tiffanymak1996

when testing ratio without the absolute value i got -1

18. godorovg

How does the value change to -1 I am lost there?

19. jeanlouie

1 is only an example here, 0.5 or any other $\epsilon < -2$ is OK, too. when $n \rightarrow \infty$, $a_{2n} \rightarrow 1$ and $a_{2n+1}\rightarrow -1$ ps:sr for my poor English

20. jeanlouie

ϵ< 2, sorry

21. tiffanymak1996

$\rho = \lim_{n \rightarrow +\infty} \frac{ (-1)^{n+1} \times e ^{\frac{ 1 }{ n }}}{ ((-1)^{n})\times e ^{\frac{ 1 }{ n }}\ }$

22. tiffanymak1996

ln both top and bottom

23. jeanlouie

yeah, ρ = -1, you got it

24. tiffanymak1996

becomes:$\lim_{n \rightarrow +\infty} -\frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } }$

25. jeanlouie

no, top or bottom may be negative, we can't use ln here.

26. tiffanymak1996

(the limit is still there but i'm just lazy: then: -n/(n+1) divide top and bottom with n: -1/(1+1/n) substitute infinity: -1/(1+0) =-1

27. godorovg

Tiffany I think you have the right answer.

28. tiffanymak1996

i do? but this is like conditionally ; yet the modal answer is divergent...

29. godorovg

Tiffany 1+1n is not -1 it is 2n

30. tiffanymak1996

model, sry

31. godorovg

2/n not -1

32. godorovg

you see where this is?

33. tiffanymak1996

in -1/(1+1/n), if i substitute n as infinity, then 1/infinity is 0, so -1/(1+0)=-1/1

34. tiffanymak1996

is this correct?

35. godorovg

when sub in n, you still have 1+1 which is 2.

36. tiffanymak1996

where does the 1+1 com from?

37. godorovg

so if we look at this -1(1+1/n) would give you add the two ones which would 2/n times that by -1 and what is the answer?

38. tiffanymak1996

so sorry, it should be -1/(1+(1/n)) not -1/(1+1/n)

39. godorovg

-1/1=-1times +1 =-1/n

40. tiffanymak1996

$\lim_{n \rightarrow +\infty}\frac{ -1 }{ 1+\frac{ 1 }{ n } }$

41. godorovg

yes..

42. tiffanymak1996

so... $\lim_{n \rightarrow +\infty} \frac{ -1 }{ 1+\frac{ 1 }{ n } }$

43. godorovg

Tiffany can the bottom be 0? or am I missing something here?

44. tiffanymak1996

the bottom can't be zero, what i did is just divide top and bottom with n

45. tiffanymak1996

and then: $\frac{ -1 }{ 1+\frac{ 1 }{ \infty } }$

46. tiffanymak1996

and since $\frac{ 1 }{ \infty }=0$,

47. tiffanymak1996

$\frac{ -1 }{ 1+0 }=-1$

48. tiffanymak1996

and i did this like 5 times and still don't get why it's divergent...

49. tiffanymak1996

T.T

50. godorovg

what happens if we -1/1+0=-1 now what if simply the bottom to -1/1=-1 than what?

51. tiffanymak1996

if it's -1 then the series is convergent, not divergent.

52. godorovg

yes..

53. tiffanymak1996

well, the model answer IS divergent.

54. godorovg

only if stays at one, -1=-1 if you add one you get 0

55. tiffanymak1996

and that's what frustrates me the most.

56. tiffanymak1996

ok...

57. godorovg

so I am taking the answer can't be 0 for the final answer?

58. tiffanymak1996

no, it's not that it can't be 0, just that how do u suddenly +1 after the limit?

59. godorovg

Because the +1 falls within the limit.

60. tiffanymak1996

oh, ok, thx, i think i get it now. :)

61. godorovg

welcome ...

62. tiffanymak1996

sry for bothering you so long...