Does the series ∑n=1∞1n2+n\sum_{n=1}^{\infty} \frac{1}{n^2 + n}∑n=1∞n2+n1 converge by the Comparison Test with bn=1n2b_n = \frac{1}{n^2}bn=n21?
Yes, because 1n2+n≤1n2\frac{1}{n^2+n} \leq \frac{1}{n^2}n2+n1≤n21 for n≥1n \geq 1n≥1 and ∑1n2\sum \frac{1}{n^2}∑n21 converges.
No, because the terms are larger than the p-series.
Yes, because the terms approach zero.
No, it diverges.