## Loser66 one year ago $$\sum_{n =1}^\infty (1/n)$$ converges. How about $$\sum_{n=1}^\infty -(1/n)$$ ? converges also? Please, help

1. Loser66

@imqwerty

2. idku

diverges!

3. Loser66

why?

4. idku

sum from 1 to ∞ of 1/n Hormonic series

5. idku

Converges means sum is defined

6. idku

diverges means the sum doesn't exist.

7. idku

In case you mixed up terms

8. idku

$$\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...$$ this is at least: $$\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...$$ $$\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...$$

9. Loser66

ok, I think I mess up something. Sorry about that .

10. ParthKohli

Harmonic series diverges. For the second, if you mean$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1-\frac{1}{2} + \frac{1}3 - \frac{1}4 + \cdots$The second sum is in fact convergent and equal to $$\ln 2$$!

11. idku

alternating series test. Converges

12. Loser66

Let me post the original one.

13. Loser66

$$\sum_{n=1}^\infty \dfrac{(-1)^n i^{n(n+1)}}{n}$$ converge or diverge?

14. idku

$$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$ it meets conditions: $$\large \displaystyle \lim_{k\to \infty} (a_n)=0$$ $$\large \displaystyle|a_n|>|a_{n+1}$$ for all n. Thus, converges.

15. idku

i mean in the limit "n -> "

16. idku

what does i do in your series?

17. idku

is that *i* to indicate imaginary value, or what>?

18. ParthKohli

it does

19. Loser66

I lost the net......:(

20. Loser66

Please, leave your guide here, I will take it if I can see what you guys write.

21. idku

$$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}$$ I haven't done series with imaginary numbers mixed in...... my guess is: $$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\le\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=\rm converges$$

22. ParthKohli

$n=1\Rightarrow i^{n(n+1)}=-1$$n=2 \Rightarrow i^{n(n+1)}=-1$$n=3 \Rightarrow i^{n(n+1)}=1$$n=4 \Rightarrow i^{n(n+1)}=1$And this continues.

23. imqwerty

$a_{n}=\frac{ -1 }{ n }$ $\lim_{x \rightarrow \infty}\frac{ -1 }{ n } =0$ so because the limit is approaching a certain value we will say that it converges

24. ParthKohli

Nah, it diverges @imqwerty

25. ParthKohli

Even though the limit of the sequence exists and is equal to zero, that is not sufficient to say that the series should converge

26. imqwerty

but whenever we take the limit and if it approaches certain value then it converges right?

27. ParthKohli

nope...

28. idku

$$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}$$ $$\large \displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)$$

29. idku

is that right?

30. ParthKohli

first of all, for convergence, the limit has to be zero and no other value. but that is not sufficient for convergence.

31. imqwerty

we will also consider the limit of partial sums right?

32. idku

yes, true dirverges test only shows divergence if lim $$\ne$$0

33. idku

divergence test ... i meant

34. ParthKohli

yes

35. idku

I was thinking it is an equivalent representation: $$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)$$

36. Empty

The divergence test is not called the convergence test for the reason that it can only show you if it diverges or not.

37. idku

yes.

38. idku

But is this reasonable? (What I said above)

39. imqwerty

why cant we say that a series diverges or converges by jst calculating the limit ?

40. Empty

The intuitive reason for what happens when you use the divergence test is this: You have an infinite sum. If the 'last' term isn't 0, then you know it diverges since infinity never ends.

41. idku

Because when the sequence an approaches 0, that can be a divergent series $$a_n|) eg Harmonic series 42. idku If I am right, \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)$$ then the series converges

43. Empty

you can have your 'last' term equal to 0 in an infinite sum, but you already went through an infinite number of terms prior to that, so those could still possibly diverge.

44. idku

and that would in fact be: ln(2)+ [ ln(2)-1 ] = 2ln(2) -1

45. idku

for my representation, if it is right

46. idku

I checked the signs, so should be fine.

47. idku

i is imaginary

48. idku

so, as parth said the i's yield: -1 -1 1 1

49. Empty

I don't understand I thought the problem was $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$

50. idku

$$\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}$$ this is the prob

51. idku

and then I suggest that just as: $$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...$$ so is: $$\displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^{n}}{n+1}\right)=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...$$

52. Empty

I was looking at the wrong problem no wonder haha

53. idku

oh no not that

54. idku

i was wrong, because my terms cancel, I have to skeep over 1

55. idku

$$\displaystyle\large\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n+1}}{2n+2}\right)$$

56. idku

this should be better: = +1/1 - 1/2 - 1/3 + 1/4 +1/5 - 1/6 - 1/7 + 1/8 +1/9 - 1/10 - 1/11 + 1/12 etc.... but that cycle is essentially just a combo of two altern. series. $$\pi/4$$ + -ln(2)/2 $$\left[\pi-2\ln2\right]/2$$

57. Empty

Actually we can probably evaluate it but to see if this series converges or diverges I think we can use a bit of a trick, look at the real and imaginary parts separately: $\Re(S) = \frac{S+S^*}{2}$$\Im(S) = \frac{S-S^*}{i2}$ These are both going to be completely real series themselves, and you can use the alternating series test to check for convergence on both: $\Re(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \cos\left( \frac{\pi}{2} n(n+1) \right)$ $\Im(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin\left( \frac{\pi}{2} n(n+1) \right)$ Inside you have $$\frac{\pi}{2}$$ so every 4 values of n you'll repeat so you could separate it out like this into 4 terms, find out what the cosine and sine terms are doing, they'll only give you 0, 1, or -1 in some funny order at worst, but we can see that the inside will repeat every 4 because of this: $\cos[ \frac{\pi}{2}n(n+1)]$ $n \to n+4$ $\cos [ \frac{\pi}{2}(n+4)(n+4+1)] = \cos[ \frac{\pi}{2}n(n+1)+\frac{\pi}{2}(4n+4n+4)]=\cos[ \frac{\pi}{2}n(n+1)]$ really it's just case $$\frac{\pi}{2}(4n+4n+4)$$ is a multiple of $$2 \pi$$. Maybe this ends up sounding more complicated than just solving this damn thing I don't know haha. But it's just 4 terms haha.

58. idku

Loser66, how are you so far?

59. Loser66

:) To me, I do something like let $$a_n = 1/n$$ $$b_n = (-1)^n i^{n(n+1)}$$ Now, just calculate $$b_n$$ Note: $$i^2 =-1\\n(n+1) is~~even$$ Hence, for n = 4k, $$b_n = (-1)^4k i^{4k (4k +1) }= 1* ((i^2)^2)^{k(4k+1)}=1$$ Hence $$b_n =1, \sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}$$. You say, is it divergent or convergent?

60. Loser66

for n= 4k+1, similar argument $b_n = (-1)^{4k+1}i^{(4k+1)(4k+2)}= -1* (i^2)^{(4k+1)(2k+1)}= -1*-1=1$ Hence $\sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}$ You say, is it divergent or convergent?

61. idku

I hope you aren't factoring all that imaginary component out of the series......

62. idku

also, it is not (-1)^(n(n+1)), but (i)^(n(n+1))

63. idku

so it is not always positive

64. idku

-1, -1, 1, 1.....

65. Loser66

Hey, all natural number can be expressed as one of the forms : 4k, 4k+1, 4k+2 , 4k+3.

66. Loser66

So as n

67. Loser66

Why do we have to do that? because that is the way we turn imaginary i to real since i^2 =-1

68. Loser66

and the circle to get real number from i is 4. I was taught that. :)

69. Loser66

$$\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong 0~~~ mod4$$ $$\sum_{n=1}^\infty a_n = -1~~ if~~ n \cong 1~~~ mod4$$ $$\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong2~~~ mod4$$ $$\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 3~~~ mod4$$ That shows the sum is divergent.