## Loser66 one year ago What is the result? (-1)^(n^2/2) = 1if n is even what is the result if n is odd? Please, help

1. Loser66

@dan815

2. DanJS

try it

3. Loser66

Can you?

4. Loser66

This is the original problem: Find the radius convergence of $$\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}$$

5. Loser66

We need an argument for i^n^2

6. ParthKohli

To be fair the root of $$-1$$ can be $$+i$$ or $$-i$$.

7. imqwerty

n=even that thing is 1 n=odd then its i

8. imqwerty

wait n=odd has got cases

9. ParthKohli

Exactly, we can't tell if it's $$i$$ or $$-i$$ right?

10. Loser66

$$i^{n^2}= (i^2)^{n^2/2}=(-1)^{n^2/2}$$

11. imqwerty

yea tru

12. Loser66

ok, then |dw:1443285237007:dw|, right?

13. DanJS

i got afraid and deleted my answer.. hah

14. ParthKohli

$\pm i$goddammit

15. DanJS

conjugate/

16. Loser66

So, what is the radius of convergence?

17. Loser66

@ganeshie8 how to argue $$(-1)^{n^2/2}$$ depend on n mod 2? the limit should be 1 and -1

18. Loser66

Let me attach my prof's arguments but I don't quite understand it. :(

19. Loser66

20. Loser66

Just similar, not my problem but the argument is the same.

21. anonymous

What do you mean by radius of convergence? That doesn't make sense to me. It's like asking the radius of convergence for $$\sum\limits_{n\ge1}\frac{1}{n}$$. Are you talking about a power series at all?

22. Loser66

The original problem is Find the radius of convergence of $$\sum_{n=0}^\infty \dfrac{z^{n^2}}{n}$$

23. Loser66

I got it =1 by finding the $$lim sup\sqrt[k]a_k=1$$

24. Loser66

The second part is :"discuss the convergence of the power series if z =i "

25. anonymous

That's much better, thanks for clarifying.

26. Loser66

Since the question is about convergence, I think we need the radius of convergence also, right? clearly, when n is even, then sum is sum of 1/n which is convergent with radius R = infinitive, right?

27. anonymous

Like you said, the series converges with radius $$1$$, which means for $$|z|<1$$. What's $$|z|$$ with $$z=i$$?

28. Loser66

1

29. Loser66

oh, divergent!! right?

30. anonymous

I would say so. I'm pretty sure a power series cannot converge for all points within its convergence disk AND a finite number of points on the boundary of the disk, but maybe I'm wrong about that... (I've never seen it happen at any rate)

31. anonymous

Hmm, looks like that might actually be possible: https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary

32. Loser66

|dw:1443289362976:dw|

33. anonymous

I don't think that's true. Consider $$\sum_{n\ge1}\frac{(-1)^n}{n}$$. You have two subsequences, $$\left\{-1,-\frac{1}{3},-\frac{1}{5},\ldots\right\}$$ and $$\left\{\frac{1}{2},\frac{1}{4},\frac{1}{6},\ldots\right\}$$, both of which diverge, yet the original series converges.

34. Loser66

But both are real, on my problem, we have only one real subsequence, the leftover is imaginary sequence

35. Loser66

The net is so bad at this end :(

36. anonymous

My point in my last comment is that, in general, you can have two series, $$\sum a_n$$ and $$\sum b_n$$ that both diverge, yet, $$\sum (a_n\pm b_n)$$ may still converge. In any case, you can split this series up case-by-case, as you/your prof have already done: \begin{align*} \sum_{n\ge1}\frac{i^{n^2}}{n}&=\sum_{k\ge1}\left(\frac{i^{(2k)^2}}{2k}+\frac{i^{(2k-1)^2}}{2k-1}\right)\\[1ex] &=\sum_{k\ge1}\frac{i^{4k^2}}{2k}+\sum_{k\ge1}\frac{i^{4k^2-4k+1}}{2k-1}\\[1ex] &=\sum_{k\ge1}\frac{(i^4)^{k^2}}{2k}+i\sum_{k\ge1}\frac{(i^4)^{k(k-1)}}{2k-1}\\[1ex] &=\sum_{k\ge1}\frac{1}{2k}+i\sum_{k\ge1}\frac{1}{2k-1} \end{align*} For this to converge, both the real and imaginary parts must converge. They do not. So yes, your reasoning works *in this particular case*.

37. Loser66

Thanks for that. I would like to make the concept clear when I replace a_n by a_k

38. Loser66

Like this $$\sum_{n =1}^\infty \dfrac{(z)^{n^2}}{n}=\sum_{k=1}^\infty a_k z^k$$ where $$a_k = 1/k ~~if~~ k = n^2$$ and $$a_k =0 ~~if~~ k\neq n^2$$

39. Loser66

Honestly, I am on and off with this. It means, I sometimes see it clearly, sometimes, not. :( I interpret the original one as = z + 0*z^2 +0*z^3 +(z^4)/2)+ 0*z^5 +0*z^6 +...+ (z^9)/3 +...

40. Loser66

Hence, if k = 1 = 1^2 , $$a_k = 1/k =1$$ if k = 2, $$a_k = 1/k =1/2$$ and this k must be a complete square of a number n =2, right? Now, confused part. if k =9, then $$a_k = 1/9$$ and n =3, but as above, $$a_k = 1/3$$ not 1/9. What am I missing?

41. anonymous

It should be $$a_kz^k=\dfrac{1}{\sqrt k}z^k=\dfrac{1}{n}z^{n^2}=a_nz^{n^2}$$.

42. Loser66

$$a_k = 1/ \sqrt k ~~ if k = n^2$$ , right?

43. Loser66

and the sum becomes $$\sum_{k=1}^\infty a_k z^k$$ and we can apply lim sup a_k, right?

44. anonymous

Yes, $$\sqrt k$$ can only be an integer if $$k$$ is a perfect square, i.e. if $$k=n^2$$ for $$n\in\mathbb{N}$$.

45. Loser66

But then, if z =-1, should I use the original one or turn it to a_k?

46. Loser66

When z =-1, the sum is $$\sum_{n=1}^\infty \dfrac{(-1)^{n^2}}{n}$$

47. Loser66

And I have to argue for its convergence also.

48. anonymous

You can use the same even/odd split to arrive at the same conclusion. You would end up with $\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}$ To use your $$n\to k$$ replacement approach, you can still make an even/odd argument: \begin{align*}\sum_{\substack{k\ge1\\k\in S}}\frac{(-1)^k}{\sqrt k}&=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{(-1)^{2p}}{\sqrt {2p}}+\frac{(-1)^{2p-1}}{\sqrt {2p-1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{1}{\sqrt {2p}}-\frac{1}{\sqrt {2p-1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\frac{\sqrt{2p-1}-\sqrt{2p}}{\sqrt {4p^2-2p}}\\[1ex] &\sim \sum_{\substack{p\ge1\\p\in S}}\frac{1}{\sqrt p}\\[1ex] &=\sum_{\substack{q\ge1\\q\in\mathbb{N}}}\frac{1}{q} \end{align*}where $$S$$ denotes the set of perfect squares.

49. Loser66

oh, I tried to put it into something totally different from yours , hihihi... but let me digest your solution first. My attempt: sum = -1 + (1/2) -(1/3) +(1/4) -(1/5) ....... and group them as $$a_n - a_{n+1}$$ where $$a_n = 1/n$$

50. Loser66

and I still stuck on the first -1.

51. Loser66

oh, I cannot do it since it is infinite series, right?

52. anonymous

Well there's nothing preventing you from doing that sort of pairing of the terms, but I'm not sure what you're trying to accomplish.

53. Loser66

One more question, how can you go from n to k at the first line of the argument?

54. Loser66

$$\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}$$

55. anonymous

That's the even-odd split.$\begin{array}{c|c|c} k&2k&2k-1\\ \hline 1&2&1\\ 2&4&3\\ 3&6&5\\\vdots&\vdots&\vdots \end{array}$

56. Loser66

so, you split it into n is even = 2k and if n is odd = 2k +1, right?

57. Loser66

I got it. Thank you so much. Thanks for being patient to me. :)

58. Loser66

But then, the odd series must be - , right?

59. anonymous

Hmm, I might have made a mistake somewhere above... If it were $$\dfrac{1}{2k}-\dfrac{1}{2k-1}$$, then the series would converge...

60. Loser66

Yes, I got $$\sum_{k=1} \dfrac{1}{4k^2 +2k}$$ then, it converges since it is < sum (1/k^2) which is convergent. Thanks again.