## idku one year ago I want to clarify something about the convergence of series and sequence.

1. idku

So, for sequence to converge 1) Bound below and above 2) Monotonic (can be always decreasing or always increasing)

2. idku

and for series: 3) The limit [n→ infinity] A(n) = 0 (am I correct)

3. zzr0ck3r

yes, but the converse does not hold on 3)

4. idku

yes, what other conditions are there?

5. idku

well, yes not on those 2 - harmonic series.

6. idku

not those 3*

7. idku

But, what else must be true?

8. zzr0ck3r

of series?

9. idku

yes convergence of series

10. zzr0ck3r

there are many many theorems

11. idku

Oh, like Σ 1/n^p then p>1 geometric series |r|<1 (and that is where ratio test is coming from) and others....

12. idku

incl alternating series test.... and others

13. idku

so it basically depends on a series in every unique cse?

14. idku

case*

15. zzr0ck3r

pretty much. there are some nice ways for some families of functions, but in general there is nothing that will deal with them all.

16. ganeshie8

if you're comfortable with sequences, then you may think of any series as a sequence of partial sums

17. zzr0ck3r

I also will assume we are talking about the real numbers with the regular topology.

18. idku

yes, that is what i am reading on wiki and other sites ganeshie

19. idku

yes. no imaginary sequences. I am not considering anything crazy

20. zzr0ck3r

Getting good with sequences is a great idea if you are planning on doing analysis. There are definitions of continuity that uses sequences that make things much nicer than dealing with the standard $$\epsilon -\delta$$ definition

21. idku

alright

22. idku

I was reading wiki just 3-5 minutes ago, and I saw that $$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{(-1)^{n+1}}{n}=\ln(2)}$$ (they don't post any link to an explanation)

23. idku

(i can create a new Q. if that is how the policy would play this time)

24. idku

(sorry if i am typing to much, don't be afraid to tell me to shut up:))

25. zzr0ck3r

This does not easily fall without other identities.

26. idku

which identities?

27. zzr0ck3r

series expansion of ln(x)

28. ganeshie8

criterian for convergence of an alternating series is an interesting one i remember seeing a very nice proof from analysis

29. zzr0ck3r

Maybe I am wrong, but I don't any basic ways to show this without more tools.

30. zzr0ck3r

You can find the taylor expansion and plug in two...

31. zzr0ck3r

But you are not proving anything....

32. idku

wait can i start from taylor series of 1/x and integrate?

33. zzr0ck3r

yeah

34. zzr0ck3r

but you already know you can do that, so what is the point :)

35. zzr0ck3r

unless you just feel like symbol pushing :)

36. ganeshie8

$\frac 1{1+x}=\sum_{n=0}^\infty (-x)^n$ integrate both sides and plugin $$x=1$$ this is fun but yeah doesn't give you the bigger picture of beatiful alternating series

37. idku

oh, we can get a series represenation for ln(x+1) :) nice

38. idku

$$\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}$$

39. ganeshie8

ln(x) has no hope at x=0 but surely it behaves well for all finite positive values of x

40. idku

oh yeah my index should have been n=0, i made that err

41. idku

and nice to hear it works for other ln values

42. idku

$$\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}$$

43. idku

oh, sorry n=0

44. idku

$$\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}$$

45. ganeshie8

now we're getting into radius of convergence and stuff for what values of x does that series converge ?

46. idku

ok, let me do ratio test:)

47. idku

|dw:1440308882162:dw|

48. idku

oh, |x|<1

49. idku

i left (-1)^n on purpose and hope correctly

50. ganeshie8

right, so its kinda tricky, the domain is restricted to [-1, 1)

51. idku

I will dare to sya in a face of god you are wrong (apologize in advance) I was just about say $$\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}$$ and because i know alternative harmonic converges, thus r=1 is valid in this case

52. ganeshie8

this is one of the most celebrated series, there are like more than 100 different proofs for its convergence... let me dig them up... :)

53. idku

we can make r=1, and just because if x=1 we get a result of "1" from the ratio test it doesn't tell us that series diverges, it is inconclusive when limit comes out with r=1, the domain is including 1.

54. idku

|dw:1440309328209:dw|this is kinda (a quite bad model) behavior of the partial sums of the alternating harmonic series

55. ganeshie8

do you mean : r = $$\color{Red}{-1}$$ is in the interval of convergence because we know that alternating harmonic series converges

56. idku

oh, x=1, not r=1 i mean

57. idku

all i was trying to say x=1 is valid, and i guess i typed much more unneccessary stuff on top

58. ganeshie8

ahh okay you have that -1 in the front..

59. idku

for interval of convergence?

60. ganeshie8

$$\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n}$$ converges for all $$x\in [-1, 1)$$

61. idku

yes, but when it is multiplied times (-1)^n then the interval is x= [-1,1] , isn't it?

62. ganeshie8

right, we're on same page :)

63. idku

oh, and then we extract that representation $$\large\color{black}{ \displaystyle \ln(\color{red}{1}+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n\color{red}{1}^{n+1}}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=1 }^{ \infty } \frac{(-1)^{n+1}}{n}}$$

64. idku

now that series makes sense ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 ...

65. idku

thank you:)

66. idku

well, should be (1)^(n$$\color{red}{-}$$1), but it is all sa,e

67. ganeshie8

looks good! its not a big deal, but below looks better : $$\large\color{black}{ \displaystyle \ln(\color{red}{x}+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n\color{red}{x}^{n+1}}{n+1}}$$ $$\large\color{black}{ \displaystyle \lim\limits_{x\to 1}~\ln(\color{red}{x}+1)=\lim\limits_{x\to 1}~\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}$$ $$\large\color{black}{ \displaystyle \ln(2)=\lim\limits_{x\to 1}~\sum_{ n=1 }^{ \infty } \frac{(-1)^{n}x^n}{n}}$$

68. idku

why is that better if I am permitted to plug in x=1 any way?

69. ganeshie8

because when deriving that series representation, before integrating, we have used geometric series restricting ourselves to $$(-1,1)$$

70. ganeshie8

so technically we're not allowed to "plugin" x=1

71. idku

oh, ok.... but we can always trick the math rules:) by the way you last line shouldn't have x^n on top, correct?

72. idku

no nvm

73. idku

well it is still same you are plugging in x=1, but you are technically not:O

74. ganeshie8

limit of an infinite sum is not same as sum of limit

75. ganeshie8

again, technically, we're not allowed to swap limit and infinite sum

76. ganeshie8

thats the reason i left it as $$x\to 1$$

77. idku

but plugging x=1 does not make the series diverge tho' the ratio test is 1 is x=1 (and jsut that no conclusions can be made, but not that it will make the sum diverge). THis series is conditionally convergent when x=1. I don't really see much problem with actually going ahead and plugging x=1. It is just illogical to me-:( :)

78. idku

the ratio test is 1, *if* x=1

79. idku

maybe i can re-express myself

80. ganeshie8

right, right, the variables are different so we can take the limit, no issues i guess

81. idku

alrighty:) I guess then I am done for now

82. idku

Will use a correct notation lim [n→∞] (thanks)^n

83. ganeshie8

lookup the proof for convergence of alternating series, its really neat... im sure you will enjoy all other proofs of various convergence tests..

84. idku

ok, lets see what they got;)

85. ganeshie8

here it is https://www.youtube.com/watch?v=tt430qiyIds&index=19&list=PL04BA7A9EB907EDAF just watch it like a movie, don't wry about all the weird terms that you might hear... :)

86. idku

oh:) It's 2:30 am in my location, i was sitting late at night because I was curious... this movie is too long for now. i have to go offline right now:) I saved the link (and if anything i can login to see the link)..... thank you very much! C(U)

87. idku

gtg, sry bye

88. ganeshie8

have good sleep :)