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# Find the radius of convergence and interval of convergence of the series.$\sum_{n = 1}^{\infty} \frac {x^n}{1 \cdot 3 \cdot 5 \cdot \cdot \cdot\cdot \cdot (2n - 1)}$

## $$I=(-\infty, \infty)$$

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case of the radius of convergence is going to be limit as n goes to infinity of absolute value of a N over and plus one where a Is this chunk here without the X value And yeah, maybe we'll well, we'll do it using the ratio test just in case this is confusing to anyone. So the ratio test would do the same type of thing Except we have the sub script with the implicit one of top now and buy this being now we do mean this whole chunk here, including the X values. Hey, so this is going to be X to the n plus one over one times three times that that that times two in minus one times two in plus one minus one. Okay, And then so that's just are being plus one. And we're dividing by being somewhat planned by the reciprocal. So now I have an X to the end over there. Enough top. We're gonna have one times three times that, that that times two and minus one. And this whole chunk here is going to cancel out with this whole chunk here and actually and plus one divided by X to the n is just going to leave us with X. So now we have limits as n goes to infinity of absolute value of X and nothing got rid of this two times in plus one minus one. So we still have that there and for using the ratio test to figure out where we get convergence. We want for this to be something less than one. But notice that this term is going to go to zero as n goes to infinity. So it doesn't matter what value we plug in for X. We're always going to end up getting zero here. Zero is certainly less than one. So X is allowed to be anything for the radius of convergence is infinity, Which is exactly what we would have gotten if we I just did this here. This will short cup would have gotten that R is equal to limit as n goes to infinity of two times in plus one minus one and that would be infinity. Okay, So if the radius of convergence is infinite than the interval of convergences, certainly going to be infinite as well

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