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If $ \sum_{n = 0}^{\infty} c_n4^n $ is convergent, can we conclude that each of the following series is convergent?

(a) $ \sum_{n = 0}^{\infty} c_n ( - 2)^n $

(b) $ \sum_{n = 0}^{\infty} c_n ( - 4)^n $

a) convergent.

b) $\left.c_{n}=(-1)^{n} /\left(n 4^{n}\right) .\right]$

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Campbell University

Baylor University

University of Michigan - Ann Arbor

if this is convergent than the ratio test is going to be able to give us some information. If the ratio test is conclusive than this would be strictly less than one. If this is convergent, but the ratio test is inconclusive, then we would be equal one. So this is all we can say is that this is less than or equal to one toe limit. As N goes to infinity of absolute value of C N plus one, overseeing is less than or equal to one fourth. There's four to the n plus one divided by foreigners for so if we divide both sides by four, then we get this inequality here. So now we can use the ratio test and part A. And with the absolute value signs, we can forget about this minus sign here So we can just treat it just like regular too, as we saw, see, the M plus one. Overseeing is less than equal to one fourth an absolute value to the M plus one over to Innes too. So this is just one half, which is less than one. So that is convergent cannon For part B. We can't conclude convergence case. That's what the problem is asking. Can we conclude convergence in the following Siri's? So, for part B, we cannot conclude convergence. So we can just used counter example here. So if CNN is minus one to the end over in times for the end. So this is what C N is equal to. We plugged that in here. Then we would get some from n equals zero to infinity. Um, so maybe we should put a ten plus one here instead of an end. So this is what we have, then some from in equals zero to infinity of CNN, times four to the end, that's going to be minus one to the end over in plus one. So this is going to converge. But notice that if we have our CNN times minus four to the end, then we would wind up with lissome, which does not converge so we can construct a si n such that this converges. But this does not converge