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Numerade Educator

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Problem 31 Hard Difficulty

If $ k $ is a positive integer, find the radius of convergence of the series

$ \sum_{n = 0}^{\infty} \frac {(n!)^k}{(kn)!} x^n $

Answer

$$R=k^{k}$$

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Video Transcript

to figure out where we get convergence here, we can use the ratio test. The Ltd's in goes to infinity of absolute value of and plus one over a M, where this whole thing here, including the X values, are they in terms actually m plus one divided by accident and it's just ex. So that's turned out to just be X times in plus one factorial to the end, plus one divided by K times in class one. Factorial. So that's our and plus one and then divided by and is multiplying by the reciprocal. So now we'LL multiply by k and factorial and we'LL divide by in factorial to the end Hey, so will lump together some of these common terms here so will write in plus one factorial to the inn plus one power as in plus one factorial to the end times in plus one factorial. And that allows us tow Get a common exponents here and then we still have one more Copy it and plus one factorial to account for. If we want to get this power of in plus one case that deals with this guy and this guy and now we have k a n factorial and we're dividing by K times in plus one factorial Kate comes in plus one. Factorial is the same thing as K n Plus Que factorial, which is the same thing as Kayin factorial times k n plus one that that that times cayenne plus Okay, all right. Barely squeezed that in there. To this Kay and Factorial will cancel out with this cayenne factorial and in plus one factorial divided by in factorial is going to just turn in tow n plus one. Okay, so we have a limit as n goes to infinity. Absolutely. How you have X times in to the end. So I guess that should be in plus one to the end. Okay, so that's this guy. And then we stuffed in plus one factorial and we're dividing by K and close want times can post two times at the thought Times can close, Kay. Okay. So we could move K copies of this in plus one. Ah, to correspond to these kay products happening down here. So you'LL see what I mean in a second. Okay, so now we have in plus one to the K that we need Thio still account for so we can have in plus one to the K happen over here. All right, so now this is limit, as in goes to infinity Absolute value of X times in plus one to the end minus K. And then I believe we stall the end plus one factorial. Yes, we d'Oh And then we can rewrite this a little bit. We can rewrite it as in plus one divided by K N plus one. We can do the same for you. The rest just put one copy of in plus one to correspond. Ah, one of the terms of the products down here. Okay, so there's Kate terms in this product here. So when we multiply, all of the numerator is we still get in plus one to the K. So where Still good. Here to remember. We're using the ratio test here. So this is going to tell us where we get convergence. I want this thing to be less than one. Okay, So as n goes to infinity in plus one over K and plus one is going to be won over K. And we have K copies of this happening. So we have one over Kay to the K. Each one of these copies is goingto have a limit Goingto won over Kay. All right, but this is just some finite number here. So as long as X is non zero, this is still going to blow up to infinity, okay? And we want for this to be less than one. So the only way that this can happen as if X is equal the zero to the interval of convergence would just be the interval on ly containing zero. It means that our radius of convergences just zero.