Have two series, just a quick check of some simple series:
$\sum _{1}^{\infty} \frac {1}{\sqrt {2n^{2}-3}}$
Considering $\frac {1}{\sqrt {2n^{2}-3}}$ > $\frac {1}{\sqrt {4n^{2}}}$ = $\frac {1}{2n}$
Since $\sum _{1}^{\infty} \frac {1}{2n}$ $\rightarrow$ Diverges, hence by the camparsion test we have that $\sum _{1}^{\infty} \frac {1}{\sqrt {2n^{2}-3}}$ diverges.
The second one is $\sum _{1}^{\infty} (-1)^{n}(1+\frac{1}{n})^{n}$
Consdering $(-1)^{n}(1+\frac{1}{n})^{n} \le |(-1)^{n}(1+\frac{1}{n})^{n}|= (1+\frac{1}{n})^{n}$
We know that $(1+\frac{1}{n})^{n}$ converges, hence we have that $\sum _{1}^{\infty}(1+\frac{1}{n})^{n}$ coverges so again by the comparison test we have that $\sum _{1}^{\infty} (-1)^{n}(1+\frac{1}{n})^{n}$ converges absolutely $\Rightarrow$ converges,
many thanks in advance.