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If $ \sum a_n $ is divergent and $ c \not= 0, $ show that $ \sum ca_n $ is divergent.

since $\sum a_{n}$ is not finite, $c \sum a_{n}$ is also not finite

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Missouri State University

Campbell University

Idaho State University

If this sum is diversion and see is non zero, let's try to show that this Siri's is also diversion. So let's argue by contradiction. So let's suppose the opposite of what we want to show the opposite of this, I suppose this converges. Then we could go ahead and write. We could go ahead and rewrite this some sense. It converges by pulling off the sea sense that some converges. This is why we're allowed to rewrite this equation. Rewrite. This is some in this form we can pull out to sea. However, if this left hand side commercial is, that means the right hand side emerges. So come on. This means that this expression on the right hand side is a real number. Let's call it X, but then we can solve for the sum of a end. So we have X oversea. But if you divide two real numbers and see is non zero, that was our assumption then. This is also remember, but this means this means that the sum of a n converges that's exactly what it means to converge anytime an infant and some serial number. That means convergence. However, we are told that the sum of the Annes diversion, and this is a contradiction. So let me highlight that contradiction using this blue color there. So the contradiction is that and converges the sum of an and that the sum of Anna's also diversion. It can't be both. Those are opposite claims. Therefore, we've reached the contradiction and our contradiction arose from this assumption that the sum cm converges. Since this assumption leads to a contradiction, it means that this statement of here this assumption cannot be true. So we must have the opposite of this. And the opposite of converging is diversion. Therefore, CNN does not converge and which we can reword that to say that is i e cnn diverges. And that's exactly what we intended to proof. So that's your final answer.