Which series converges according to the basic comparison test with ∑1n2\sum \frac{1}{n^2}∑n21?
∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{n^2+1}∑n=1∞n2+11
∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}∑n=1∞n1
∑n=1∞n2n2+1\sum_{n=1}^{\infty} \frac{n^2}{n^2+1}∑n=1∞n2+1n2
∑n=1∞1n−1\sum_{n=1}^{\infty} \frac{1}{n-1}∑n=1∞n−11