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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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Missouri State University

Harvey Mudd College

University of Nottingham

Boston College

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.

University of Illinois at Urbana-Champaign