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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 ? } $

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    $\begingroup$ The comparison test does not work that way. If the bigger series is divergent, you cannot conclude that the smaller series is also divergent. Only if the smaller one is divergent, then you can say the bigger one is divergent. So your $p$-test argument is the way to go here. $\endgroup$ – Nick Mar 22 '16 at 21:40
  • $\begingroup$ What are those integrals doing there? $\endgroup$ – zhw. Mar 22 '16 at 22:42
  • $\begingroup$ Ok so can I reverse it and say that because 1/n^2 converges and 1/n! is less than 1/n^2 it must also converge ? $\endgroup$ – Doug Ray Mar 22 '16 at 23:06
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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.

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