Convergence of (a)$\sum{\frac{1}{(\ln(n))^n}}$ (b)$\sum({\frac{2}{(-1)^n-3}})^n$ (c) $\sum{\frac{1}{\sqrt{n}\ln(n)}}$ For (a)$\sum{\frac{1}{(\ln(n))^n}}$
$\ln(n)>2$       $ \forall n > e^2$
$(\ln(n))^n>2^n$  $ \forall n > e^2$
$\frac{1}{(\ln(n))^{n}}<\frac{1}{2^n}$  $ \forall n > e^2$
$\sum{\frac{1}{2^n}}$ is convergent by geometric series test
$\sum{\frac{1}{(\ln(n))^n}}$ is convergent. 
Is this argument correct?
for (b)
$\sum({\frac{2}{(-1)^n-3}})^n$
On using the root test I'm left with 
$\sum{\frac{2}{|(-1)^n-3|}}$
this gives no definite information..how can I proceed?
for (c) $\sum{\frac{1}{\sqrt{n}\ln(n)}}$
I am unable to use comparison test, can't use divergence or integral test as well or root or ratio.
Please help.
 A: Your answer to $(a)$ is correct.
Concerning the series $(b)$ one may consider the partial sums, for $N\ge1$,
$$
\begin{align}
S_{2N}:=\sum_{n=0}^{2N}\left({\frac{2}{(-1)^n-3}}\right)^n=&\sum_{p=0}^N\left({\frac{2}{1-3}}\right)^{2p}+\sum_{p=1}^N\left({\frac{2}{-1-3}}\right)^{2p-1}
\\\\=&\sum_{p=0}^N1-\frac{2}{3}+\frac2{3 \cdot4^{N}}
\\\\=&N+\frac{1}{3}+\frac2{3 \cdot4^{N}}
\end{align}
$$ giving that $S_{2N} \to \infty$ as $N \to \infty$ yielding the divergence of the given series.
The series $(c)$ may be seen to diverge by the comparison test since, as $N \to \infty$,
$$
\sum_{n=2}^N \frac{1}{\sqrt{n}\ln(n)} \ge\int_2^N \frac{dx}{\sqrt{x}\ln(x)}\ge\int_2^N \frac{dx}{x\ln(x)}=\ln \left(\ln N\right)-\ln \left(\ln 2\right) \to \infty.
$$
A: Your reasoning for (a) is perfect.
For (b), note that the terms do not approach zero since each even term is equal to $1$. Thus it diverges by the divergence test.
For (c), note that for large enough $n$ (say $n \ge N$ where $N$ is some fixed natural number), we have $\ln(n) \le n^{1/4}$ and thus $$\frac{1}{\sqrt n \ln(n)} \ge \frac{1}{n^{3/4}}.$$ Hence $$\sum_{n\ge N} \frac{1}{\sqrt n \ln(n)} \ge \sum_{n \ge N} \frac{1}{n^{3/4}}.$$ The latter diverges since it is a $p$-series with $p \le 1$. Thus the former diverges as well.
