Find the radii of convergence of the following series. I'm having trouble with c) and d) I tried using the ratio test on d but it got pretty messy.

 A: I'll use the root test for $(c)$ which gives the radius of convergence 

$$ R = \frac{1}{ \limsup_{n\to \infty} |a_n|^{1/n}} = \frac{1}{4} .$$

A: For (d),
let's try separating the
main term:
$\frac{(-1)^n}{n-(-1)^n\sqrt{n}}-\frac{(-1)^n}{n}
=(-1)^n\frac{n-(n-(-1)^n\sqrt{n})}{(n-(-1)^n\sqrt{n})(n)}
=(-1)^n\frac{(-1)^n\sqrt{n}}{(n-(-1)^n\sqrt{n})(n)}
=\frac{\sqrt{n}}{(n-(-1)^n\sqrt{n})(n)}
=\frac{1}{(n-(-1)^n\sqrt{n})\sqrt{n}}
$.
Therefore,
$\begin{array}\\
\sum_{n=1}^M \frac{(-1)^n}{n-(-1)^n\sqrt{n}}
&=\sum_{n=1}^M \left(\frac{1}{(n-(-1)^n\sqrt{n})\sqrt{n}}+\frac{(-1)^n}{n}\right)\\
&=\sum_{n=1}^M \frac{1}{(n-(-1)^n\sqrt{n})\sqrt{n}}+\sum_{n=1}^M\frac{(-1)^n}{n}
\\
\end{array}
$
The first series converges
absolutely
(by comparison with
$\frac1{n^{3/2}}$)
and the second is an alternating series
of decreasing terms,
so it converges conditionally.
Therefore
their sum converges
conditionally.
Note that if the terms are
$\frac{(-1)^n}{n-(-1)^nn^a}$
where
$a < 1$,
the difference is
$\frac{1}{(n-(-1)^n\sqrt{n})n^{1-a}}$,
so the same result applies,
by the first series
compared to
$\frac1{n^{2-a}}$,
since
$2-a = 1+(1-a) > 1$.
