Identify which series converges by the Direct Comparison Test with ∑1n2\sum \frac{1}{n^2}∑n21.
∑n=1∞1n2+n\sum_{n=1}^{\infty} \frac{1}{n^2 + n}∑n=1∞n2+n1
∑n=1∞nn2+1\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}∑n=1∞n2+1n
∑n=1∞n+1n2−0.5\sum_{n=1}^{\infty} \frac{n+1}{n^2 - 0.5}∑n=1∞n2−0.5n+1
∑n=1∞n2n2+100\sum_{n=1}^{\infty} \frac{n^2}{n^2+100}∑n=1∞n2+100n2