Answer
The first five terms are: $1, -0.25 , 0.11111111, -0.0625, 0.04$
This converges to $0$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $\dfrac{{(-1)^{n+1}}}{{n^2}}$.
$\implies n=1$: $\dfrac{{(-1)^{1+1}}}{{1^2}} =1$
$\implies n=2$: $\dfrac{{(-1)^{2+1}}}{{2^2}}=-0.25$
$\implies n=3$: $\dfrac{{(-1)^{3+1}}}{{3^2}}=0.11111111$
$\implies n=4$: $\dfrac{{(-1)^{4+1}}}{{4^2}}=-0.0625$
$\implies n=5$: $\dfrac{{(-1)^{5+1}}}{{5^2}}=0.04$
We see that $\dfrac{-1}{n^2} \leq \dfrac{(-1)^n}{n^2} \leq \dfrac{1}{n^2}$
But $\lim\limits_{n \to 0} \dfrac{-1}{n^2}=0$ and $\lim\limits_{n \to 0} \dfrac{1}{n^2}=0$
Therefore, $\lim\limits_{n \to 0} \dfrac{(-1)^n}{n^2}=0$ by the squeeze theorem and converges to $0$.
