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Find the radius of convergence and interval of convergence of the series.

$ \sum_{n = 1}^{\infty} \frac {( - 1)^n 4^n}{\sqrt{n}} x^n $

$$\left(-\frac{1}{4}, \frac{1}{4}\right]$$

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Missouri State University

University of Michigan - Ann Arbor

Idaho State University

cave. So are, as usual, is limit as and goes to infinity Hey, and over a n plus one and should be opposite values here. That's the value is going to get rid of this minus one to the end because the ends here by a and we mean minus one of the end times foreign divided by squared of end. So these guys here, those are aliens. So this is going to turn out to be limit as n goes to infinity of four to the end over square root of n and then dividing by and plus one is the same thing as multiplying by the reciprocal. So we get square root of n plus one divided by four to the end, plus one for the end of that of about four to the M plus one is just one over four. We have one fourth, and then we can combine these square roots. Limn is n goes to infinity of n plus one over n is one. So this guy turns into squared of one, which is just one, and we're multiplying that by one fourth. So we get one fourth as the radius of convergence for the interval of convergence. We just need to figure out whether or not we get convergence that one fourth and whether or not we get convergence at minus one fourth. Okay, so when exes minus one fourth, be plugging minus one fourth into here, what you're going to end up getting is some from in equals one to infinity of minus one to the end for to the end, over square root of n and then times minus one over four to the end. But remember, minus one to the end times for the end is the same thing as minus four to the end. So minus four to the end, times minus one fourth to the end, those will cancel out. So this is just going to be one over the square root of n square. Root of n is just like in to the one half power. And when that exponents is something that's less than one or equal, the one we get divergent. So this is going to diverge. And when X is equal to positive one fourth If we were to plug that into here would get the same kind of thing here, except we'd be alternating sign. So if we're alternating sign, the alternating signed test would give us convergence here because these terms are going to go to zero. So here, this is divergent. And I guess I'll just write this on another page. If X is equal to positive one fourth you're gonna end up with this alternating signed test tells us that this is going to converge. So minus one fourth we have to throw out because that would give us divergence. One fourth gives us convergence, so we include that, and this is our interval of convergence here.