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try it

Can you?

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.

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.

1

oh, divergent!! right?

|dw:1443289362976:dw|

The net is so bad at this end :(

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

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?

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.

and I still stuck on the first -1.

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

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

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?