Problem 14 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {x^{2n}}{n!} $

Answer

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Video Transcript

since we haven't X to the two and here instead of just an axe to the end, we needto start from the top when we're figuring out the Raiders of Convergence. So as long as you remember what the radius of convergence means and we should be fine and just use the ratio test to proceed the ratio test we look at limit as n goes to infinity of a and plus one divided by a n Now, when we say and we mean this whole thing, including the X values on previous problems, we just meant you know, this whole thing except for the X values that here a n is X to the two and divided by in factorial. Okay, so this turns into limit as n goes to infinity of absolute value of X to the two times in plus one divided by in plus one factorial That's a and plus one. And we're dividing by a m. So we're multiplying by the reciprocal. So we're multiplying by in factorial divided by X to the two n So in factorial is one times two times three times dot, dot dot times in. So if you divide that by N Plus one factorial There's going to be a lot of cancelling that will occur and we'll just be left with in Plus one and then we have X to the two and plus one divided by extra to end so X to the two times in plus one is X to the two n plus two so excellent to and we'LL cancel out with this x to the two n and we'LL just have X squared here. And when you're doing the ratio test you want for this to be less than one. So figure out which values of X is going to make this less than one. Well, all of the values of acts are going to make this less than one. Doesn't matter if X is five ten a million, all all the values of actually going to make this less than one. So our radius of convergence is actually infinity and the interval of convergence is from minus infinity to infinity. Now, if it didn't happen to work out this way, the way you find the radius of convergence, if you only know the interval of convergence is toh, take the length of that interval and then divide it by two. So if this was from minus two, positive to than the radius of convergence would just be, too. In this case, it happens to be infinite and this ratio test here. Remember that in the case where you get equal toe one, you need to be careful when it's equal to one. That test is inconclusive, and you just need to figure out whether or not you get convergence or divergence another way either limit comparison test or a regular comparison test. But when it's equal the one, the test is not conclusive. But this is, Ah, good idea for just figuring out the radius of convergence. And in this case, the radius of convergence is infinity and the interval of convergences minus infinity to infinity. So we get convergence everywhere here.