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

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

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This is the original problem: Find the radius convergence of \(\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}\)
We need an argument for i^n^2
To be fair the root of \(-1\) can be \(+i\) or \(-i\).
n=even that thing is 1 n=odd then its i
wait n=odd has got cases
Exactly, we can't tell if it's \(i\) or \(-i\) right?
\(i^{n^2}= (i^2)^{n^2/2}=(-1)^{n^2/2}\)
yea tru
ok, then |dw:1443285237007:dw|, right?
i got afraid and deleted my answer.. hah
\[\pm i \]goddammit
conjugate/
So, what is the radius of convergence?
@ganeshie8 how to argue \( (-1)^{n^2/2}\) depend on n mod 2? the limit should be 1 and -1
Let me attach my prof's arguments but I don't quite understand it. :(
Just similar, not my problem but the argument is the same.
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?
The original problem is Find the radius of convergence of \(\sum_{n=0}^\infty \dfrac{z^{n^2}}{n}\)
I got it =1 by finding the \( lim sup\sqrt[k]a_k=1\)
The second part is :"discuss the convergence of the power series if z =i "
That's much better, thanks for clarifying.
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?
Like you said, the series converges with radius \(1\), which means for \(|z|<1\). What's \(|z|\) with \(z=i\)?
1
oh, divergent!! right?
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)
Hmm, looks like that might actually be possible: https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary
|dw:1443289362976:dw|
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.
But both are real, on my problem, we have only one real subsequence, the leftover is imaginary sequence
The net is so bad at this end :(
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*.
Thanks for that. I would like to make the concept clear when I replace a_n by a_k
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 \(a_k =0 ~~if~~ k\neq n^2\)
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 +...
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?
It should be \(a_kz^k=\dfrac{1}{\sqrt k}z^k=\dfrac{1}{n}z^{n^2}=a_nz^{n^2}\).
\(a_k = 1/ \sqrt k ~~ if k = n^2\) , right?
and the sum becomes \(\sum_{k=1}^\infty a_k z^k\) and we can apply lim sup a_k, right?
Yes, \(\sqrt k\) can only be an integer if \(k\) is a perfect square, i.e. if \(k=n^2\) for \(n\in\mathbb{N}\).
But then, if z =-1, should I use the original one or turn it to a_k?
When z =-1, the sum is \(\sum_{n=1}^\infty \dfrac{(-1)^{n^2}}{n}\)
And I have to argue for its convergence also.
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.
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 \(a_n = 1/n\)
and I still stuck on the first -1.
oh, I cannot do it since it is infinite series, right?
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.
One more question, how can you go from n to k at the first line of the argument?
\(\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}\)
That's the even-odd split.\[\begin{array}{c|c|c} k&2k&2k-1\\ \hline 1&2&1\\ 2&4&3\\ 3&6&5\\\vdots&\vdots&\vdots \end{array}\]
so, you split it into n is even = 2k and if n is odd = 2k +1, right?
I got it. Thank you so much. Thanks for being patient to me. :)
But then, the odd series must be - , right?
Hmm, I might have made a mistake somewhere above... If it were \(\dfrac{1}{2k}-\dfrac{1}{2k-1}\), then the series would converge...
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.

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