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Numerade Educator



Problem 2 Medium Difficulty

Suppose $ \sum a_n $ and $ \sum b_n $ are series with positive terms and $ \sum b_n $ is known to be divergent.
(a) If $ a_n > b_n $ for all $ n, $ what can you say about $ \sum a_n? $ Why?
(b) If $ a_n < b_n $ for all $ n, $ what can you say about $ \sum a_n? $ Why?


a. then $\sum a_{n}$ is divergent $[$ This is part (ii) of the comparison test $]$
b. Click to see


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Video Transcript

So we have a N and being a series with pub terms and beings known to divergence. So if a and it's great and being for each term look at me say about eight and and why? So for this question, we can conclude that the series A n well, some of to infinity if we sum up and over the natural number because, uh, we know that we know that for each term for each end, we have a m greater, then be in. So we have this inequality. If we assemble, we're in, and here we can conclude that A and s diverted. But for question be can we conclude that a is convergent? Actually, not because, uh, be an establishment. That's correct. But a is less than B in. We do not have enough clue to say am is convergent or divergent. Um, I can share some examples. Let's say if being equals the harmonic series one over in and A N is one or two in, so we know that b and it's going to be divergence and am is just one half beaten. It's also divergent, but if we substitute a M by let's see into that equals one or elsewhere. And this is definitely less than or equal to one or end. And now the serious A until that is going to be, can work. It's going to convergent. So this is less than infinity. What? Less than infinity. Okay, we start already here.