For what values of the alternating series is convergent?
Answer. It is convergent for all .
Solution. Due to Leibniz rule, it suffices to prove that is decreasing for all and all since for any .
Therefore,
It is clear that the latter term in the above resulting is eventually negative since and .
Hence is decreasing for all sufficiently large and all positive, and this implies that the series is convergent for all .
correct! using limit to prove that a function is eventually negative was the idea of this problem! many calculus I students probably don't know this important point.
you didn't need the inequality though because you could have also argued that:
for because
the reason that i assumed is that the series is obviously "absolutely" convergent for
I was now thinking about it!
But i guess a comparison can be given in this way,
if the series formed by the minorants of positive terms and majorants of negative terms converges, then the first one also converges.
For instance, rewrite the series in the following form
then find two positive sequences such that and for all sufficiently large, and converges.