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Find the interval of convergence of each of the following power series; be sure t0 investigate the endpoints of the interval in each case.X(-1)"z"2 20 247...

Question

Find the interval of convergence of each of the following power series; be sure t0 investigate the endpoints of the interval in each case.X(-1)"z"2 20 247 11. 246"9 (-"x" (Zn)!22 10. 28 (2n)3/2Ee-1"r"2" 12_ En(-Zc)"

Find the interval of convergence of each of the following power series; be sure t0 investigate the endpoints of the interval in each case. X(-1)"z" 2 20 247 11. 246" 9 (-"x" (Zn)! 2 2 10. 28 (2n)3/2 Ee-1"r"2" 12_ En(-Zc)"



Answers

Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=1}^{\infty}\left[\frac{2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2 n}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot(2 n+1)}\right] x^{2 n+1}$$

Middle of conversions for this power series do that use the ratio test first taking the limit as n approaches infinity of two x to the this one By two x 2 X. to the power here. This gives us two x times the value of two X is less than one year. So then got that supply of X is less than white house. So then we have from negative one half To 1/2 as a potential. We need to test the endpoints so let's go ahead and try that up. We could see at X equals negative one half. We get -1 to the head power which the bridges then at X ankles one half, get onto the and power it's also and I purchase so therefore her interval convergence from negative one half, 2.5.

Several of convergence first were introduced. Take the limit as N approaches infinity, which is the ratio toes apply that to this. Here's with M plus one factorial over Times I texted them plus one. all that over two and plus two factorial times the reciprocal of the original. Just two and tutorial divided by N. Factorial X at the end. Alright, so simplifying this out here on the top, we have 10 plus one time sex On the bottom, we have two, n plus two Times 2 & Plus one. Okay, not taking the limit as N approaches infinity here, you see that there's two ends on the bottom ah components here. So it's like N squared and put it on top so this will go to zero, which is always less than one. Therefore the series converges from negative infinity to infinity.


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