## tiffanymak1996 Group Title is this absolutely convergent or conditionally or divergent? 2 years ago 2 years ago

1. tiffanymak1996 Group Title

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

2. Herp_Derp Group Title

Divergent...

3. tiffanymak1996 Group Title

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

4. Herp_Derp Group Title

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 Group Title

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

6. godorovg Group Title

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

7. jeanlouie Group Title

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 Group Title

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

9. tiffanymak1996 Group Title

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 Group Title

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

11. godorovg Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

16. godorovg Group Title

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

17. tiffanymak1996 Group Title

when testing ratio without the absolute value i got -1

18. godorovg Group Title

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

19. jeanlouie Group Title

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 Group Title

ϵ< 2, sorry

21. tiffanymak1996 Group Title

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

22. tiffanymak1996 Group Title

ln both top and bottom

23. jeanlouie Group Title

yeah, ρ = -1, you got it

24. tiffanymak1996 Group Title

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

25. jeanlouie Group Title

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

26. tiffanymak1996 Group Title

(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 Group Title

Tiffany I think you have the right answer.

28. tiffanymak1996 Group Title

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

29. godorovg Group Title

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

30. tiffanymak1996 Group Title

model, sry

31. godorovg Group Title

2/n not -1

32. godorovg Group Title

you see where this is?

33. tiffanymak1996 Group Title

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

34. tiffanymak1996 Group Title

is this correct?

35. godorovg Group Title

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

36. tiffanymak1996 Group Title

where does the 1+1 com from?

37. godorovg Group Title

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 Group Title

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

39. godorovg Group Title

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

40. tiffanymak1996 Group Title

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

41. godorovg Group Title

yes..

42. tiffanymak1996 Group Title

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

43. godorovg Group Title

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

44. tiffanymak1996 Group Title

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

45. tiffanymak1996 Group Title

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

46. tiffanymak1996 Group Title

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

47. tiffanymak1996 Group Title

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

48. tiffanymak1996 Group Title

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

49. tiffanymak1996 Group Title

T.T

50. godorovg Group Title

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

51. tiffanymak1996 Group Title

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

52. godorovg Group Title

yes..

53. tiffanymak1996 Group Title

well, the model answer IS divergent.

54. godorovg Group Title

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

55. tiffanymak1996 Group Title

and that's what frustrates me the most.

56. tiffanymak1996 Group Title

ok...

57. godorovg Group Title

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

58. tiffanymak1996 Group Title

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

59. godorovg Group Title

Because the +1 falls within the limit.

60. tiffanymak1996 Group Title

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

61. godorovg Group Title

welcome ...

62. tiffanymak1996 Group Title

sry for bothering you so long...