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Numerade Educator



Problem 43 Easy Difficulty

Show that if $ a_n > 0 $ and $ \lim_{n \to \infty} na_n \not= 0, $ then $ \sum a_n $ is divergent.


$\sum a_{n}$ must be divergent.


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Video Transcript

So let's suppose that the limit here and a and A this some number l and were given that it is non zero since the limit evils l This tells us there exist and capital end such that if we take a little and to be larger than this bigon, then the absolute value of any end minus l is less than l over to. So here let's let's note that l is a positive number because it's coming. This limit is just multiplying too. Positive numbers. We're giving the ans positives. So this implies and a and minus l is bigger than over to you. And I could solve this for an now. This tells us that the sum of the AI ends is larger than l or to let me write this one over end. But this series on the right diverges this is just the harmonic series. Or you could even just use the pee test with t equals one. So diverges therefore, by the comparison test this larger Siri's on the left side must also converge. Excuse me. Also diverges No. So we have We just proved that it diverges and that's the final answer