Convergence of the series below $$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}$$
I did:
$$\lim_{n\to \infty}\Biggr\vert\frac{(-1)^n}{\sqrt{n}}\Biggr\vert$$
$$\lim_{n\to \infty}\frac{1}{n^\frac{1}{2}}=0<1$$
So diverges by the Ratio Test, right?
And have this series:
$$\sum_{n=1}^\infty \bigg(\frac{3n^3+2}{2n^4+1}\bigg)^n$$
I solved and found:
$$\lim_{n\to \infty}\frac{n^4(\frac{3}{n}+\frac{2}{n^4})}{n^4(2+\frac{1}{n^4})}=0<1$$
So is absolutely convergent by the Root Test, right?
But the answers options are:
A)Conditionally convergent and absolutely convergent.
B)Both are absolutely convergent.
C)Both are divergent.
D)Conditionally convergent and divergent.
None matches with my answers. Where am I going wrong?

As for your first series, you did not show it diverges, but you can show (compare it with $1/\sqrt{n}$) that the series of the absolute values diverges, i.e. your series is not absolutely convergent. Using Leibniz's criteria for alternating sign series you can show it converges, making it simply (or conditionally) convergent. You are correct with respect to the second series, assuming that your calculations indicate you understood the series is bounded by a convergent geometric series.
