For which of the following series does thealternating testensureconvergence?

\displaystyle \sum_{n=0}^\infty (-1)^n \frac{(n^2)!}{(2n)!}n=0(1)n(2n)!(n2)!

\displaystyle \sum_{n=1}^\infty (-1)^n \ln \frac{n+1}{n}n=1(1)nlnnn+1

\displaystyle \sum_{n=1}^\infty (-1)^n \arctan (\pi n)n=1(1)narctan(n)

\displaystyle \sum_{n=0}^\infty (-1)^nn=0(1)n

\displaystyle \sum_{n=0}^\infty (-1)^n\left(\frac{-1}{5} ight)^nn=0(1)n(51)n

\displaystyle \sum_{n=1}^\infty (-1)^n \frac{2n-1}{2n+1}n=1(1)n2n+12n1

\displaystyle \sum_{n=1}^\infty (-1)^n \ln^2 \left( \cos \frac{1}{n} ight)n=1(1)nln2(cosn1)

\displaystyle \sum_{n=1}^\infty (-1)^n \arctan\frac{1}{n}n=1(1)narctann1

\displaystyle \sum_{n=1}^\infty (-1)^n \left( \frac{3n-1}{n^2} ight)^nn=1(1)n(n23n1)n

\displaystyle \sum_{n=0}^\infty (-1)^n \frac{(n!)^2}{(2n)!}n=0(1)n(2n)!(n!)2