Decide whether the following series converge or diverge; in the case of convergence, indicate whether it is absolute: (i) $\sum_{n=1}^{\infty} n^{-1/2}i^n$ ; (ii) $\sum_{n=2}^{\infty}n^{-1}[Log(n+iy)]^{-2}$ ; (iii) $\sum_{n=0}^{\infty}(n!)^2[(2n)!]^{-1}(3+4i)^n$ ; (iv) $\sum_{n=-\infty}^{\infty}2^n\sec(ni)$ ; (v) $\sum_{n=1}^{\infty}(1+n^{-1})^{-n^2}(1+2i)^ne^{ni}$.
(i) $\lim_{n\rightarrow \infty}|n^{-1/2}i^n|=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0$ so, I can not decide if the series converges or diverges by this method, note that $\sum_{n=1}^{\infty} |n^{-1/2}i^n|=\sum_{n=1}^{\infty} n^{-1/2}$ is a series p where $p=1/2$ and so I could say that the series diverges?
(ii) $\lim_{n\rightarrow \infty}|n^{-1}[Log(n+iy)]^{-2}|=0$ so, I can not decide if the series converges or diverges by this method
How can I determine if these series diverge or converge in an efficient and easy way? Thank you very much
(i) With Raabe's test it's converge $$\lim_{n\to\infty} n\left(1-\Big|\dfrac{a_{n+1}}{a_n}\Big|\right)=\lim_{n\to\infty} n\left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)<\dfrac12$$
(v) With Root's test it's converge $$\lim_{n\to\infty} \sqrt[n]{\Big|\left(1+\dfrac1n\right)^{-n^2}(1+2i)^ne^{ni}\Big|}=\dfrac{\sqrt{5}}{e}<1$$