Originally Posted by

**ThePerfectHacker** You want to prove $\displaystyle a_n \to 1$.

Thus, we have to show $\displaystyle \left| \frac{n^2+(-1)^n}{(n+1)^2} - 1 \right| < \epsilon$.

Thus, $\displaystyle \left| \frac{n^2 + (-1)^n - (n+1)^2}{(n+1)^2}\right| < \epsilon$.

Thus, $\displaystyle \left| \frac{n^2 + (-1)^n - n^2 - 2n - 1}{(n+1)^2} \right| < \epsilon \implies \left| \frac{(-1)^n - 2n}{(n+1)^2} \right| < \epsilon$.

But note, $\displaystyle \left| \frac{(-1)^n - 2n }{(n+1)^2} \right| \leq \frac{2n+1}{n^2} \leq \frac{2n + n}{n^2} \leq \frac{3}{n}$.

So if we can make $\displaystyle 3/n < \epsilon$ then $\displaystyle |a_n - 1| < \epsilon$. And this of course can be made by choosing $\displaystyle N > 3/\epsilon$.