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Problem 13 Medium Difficulty

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

$ \sum_{n = 1}^{\infty} \frac {n}{2^n(n^2 + 1)} x^n $

Answer

$$[-2,2)$$

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Video Transcript

the radius of convergence. Here is our limit as n goes to infinity of absolute value of a n over n plus one where AM is in over to the n times n squared plus one. So this is limit as n goes to infinity of n over to the end Times n squared plus one and then dividing by a seven plus one is the same thing as multiplying by the reciprocal. So multiply by two to the end times n plus one quantity squared plus one and then dividing by n plus one. So that's limit as n goes to infinity of to the n plus one divided by two to the end is just too And then we can group this in and this n plus one together just to make it a little easier to look at And then we can We're left with n plus one squared plus one divided by and squared plus one. So up here this is a degree to polynomial with leading coefficient one down here this is a degree to polynomial with leading coefficient one. So as n goes to infinity, this is going to go to one. And as n goes to infinity and over n plus one is also going to go to one. So this turns into two times one times one, which is just to that's the radius of convergence. For the interval of convergence, we need to figure out whether or not we include X equals minus two and whether or not we include X equals two. So if X is equal to minus two, there are some turns into some from n equals, one to infinity of and and then we have minus two to the N divided by two to the end. The two Italians will cancel. Then we'll just have this minus one to the end. Here, this n squared plus one down here. So now we're alternating sign if we just plug in X equals minus two here, and these terms are going to be going to zero. So the alternating sign test is going to give us convergence here. So now we need to figure out what happens when X is equal to two. Okay, if we plug in X equals to hear this to the end will cancel out with this to the end, and then we're just going to have in over n squared plus one. Okay, so call these terms are be in terms now, I guess we can we just call them are? Yeah, sure. We'll call him be in terms, and then we're gonna do the limit comparison tests. We're gonna want to compare this to something else. So for the limit comparison test, you want to think about what happens as n goes to infinity. So as n goes to infinity, this plus one is negligible. So the terms will compare it to will just be in, over in squared, and we'll pass out this plus one. But that's going to be the same thing as one over in. Okay, So by construction, we set it up so that we have limit as n goes to infinity. Absolute value of being overseeing is one for the limit comparison test. As long as you get something that's finite and non zero here, then whatever happens when we sum up, these see in terms will get the same behavior as when we sum up. These be in terms summing up these see in terms we get divergence because that's the harmonic series. Therefore, we get divergence when we sum up, these be in terms, so X equals two. We have to throw out of our interval of convergence as we mentioned, minus two works and by the limit comparison tests we see that too does not work. So this is our interval of convergence here.

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