Discuss convergence of $\sum_{n=1}^\infty \frac{i^{n^2}}{n}$. I had to show that the radius convergence of the power series $\sum_{n=1}^\infty \frac{z^{n^2}}{n}$ is $1$, and I have it done. The second part is discussing the  convergence if $z =i$,  that is $$\sum_{n=1}^\infty \frac{i^{n^2}}{n}.$$ 
My attempt: $i^{n^2}=(i^2)^{n^2/2}=(-1)^{n^2/2}$ depend on $n \pmod 2$ 
If $n$ is even, it is  $1$ and the sum is convergent. But I do not know how to argue if $n$ is odd, the value of $i^{n^2/2}= i^{(2k+1)^2/2} = \pm i$.
Please help me. 

As you remarked correctly when $n$ is even then $n^2$ is a multiple of $4$ and does $i^{n^2}$ is $1$. The series of the even terms is thus $$\sum_{n\in \mathbb{N}, \text{even}}  \frac{1}{n}= \sum_{m=1}^{\infty} \frac{1}{2m} = \frac{1}{2}\sum_{m=1}^{\infty} \frac{1}{m} $$
and this diverges contrary to what you say. 
This does not yet mean that the complete series diverges. But coupling the above observation with your observation that the odd terms will be purely imaginary, you can conclude that the series diverges. 
To be detailed, consider real part of each partial sum. This is the sum of the even terms in each partial sum, and this diverges. Thus, the sequence of partial sums can never converge as its real part diverges. 
Side-note: for $n$ odd, one has that $n^2$ always leaves remainder $1$ on division by $4$ so for odd $n$ one has $i^{n^2}= i$, but this is not crucial for the argument.    
