I was given this problem:
$\sum_{n=1}^\infty \frac{1}{n!} \\ \text{By the comparison test :} \\ \text{If} \sum_k^{\infty} a_n <= \sum_k^{\infty} b_n \text{ Then} \text{ if } b_n \text{ diverges so does } a_n. \\ \text{At this point my first thought was to compare it to } 1/n! \text{ since I already know its behavior.} \\ \text{So since } \frac{1}{n!} < \frac{1}{n} \text{ and } \frac{1}{n} \text{ is the Harmonic series. } b_n \text{ is divergent and therefore } a_n \text{ is divergent.}\\ \text{However the same proof could be set up for } \frac{1}{n^2} \text{ and } \frac{1}{n^2} \text{ since} \frac{1}{n^2} \text{ is a convergent p-series with } n > 1 \\ \text{So which of these are correct ? } $
Actually, you have it backwards. As Nick has noted, you can only conclude that $b_n$ will diverge if $a_n$ diverges, not the other way around.