Your challenge tonight is to come up with a series that can be shown to converge using (either version of) the comparison test. We’ll vote on the best example tomorrow. Here’s my entry.
\[\sum_{n=1}^\infty \frac{1}{2^n-n^2}\]
Dr. McDonald's Mathematics Course Blog
This is my entry:
$$\sum_{n=1}^∞ \frac{1}{\ln(2^n )-n^2}$$
mine is: \[\sum_{n=1}^{\infty}\frac{\left ( 2 + n^{3} \right )}{2^{n} \left ( n^{8}\left ( 9n^{2} + e^{4}\right ) \right )}\]
Here’s Nowshin’s entry:
\[\sum_{n=1}^\infty \frac{\sqrt{3n^2+5n+2}}{n^3+4}\]