Determine the convergence of ∑n=1∞1+(−1)n⋅nn2\sum_{n=1}^{\infty} \frac{1 + (-1)^n \cdot n}{n^2}∑n=1∞n21+(−1)n⋅n.
When nnn is even: term =1−nn2= \frac{1-n}{n^2}=n21−n; when nnn is odd: term =1+nn2= \frac{1+n}{n^2}=n21+n. The series diverges due to harmonic-like components
The series converges absolutely
The series diverges by the Divergence Test
The series converges because all terms are bounded