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

## $$[-1,1)$$

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Okay. The radius of convergence are is just the limit, as in goes to infinity of the absolute value of a N over A and plus one where one over two and minus one is going to be our way in term. So this is Lim has n goes to infinity of two times in plus one minus one over two and minus one because dividing by something is the same thing as multiplying by the reciprocal. So that's how this two times in plus one minus one ended up being on top here. Okay, so this is something that's linear on top, something that's linear on the bottom. And the leading coefficient is too in both these cases. So this limit just turned out to be one. Okay, And now we just need to figure out for the interval of convergence, whether or not one in minus one get included. So when acts is equal to minus one, we need to figure out what happens next is minus one. We have minus one to the end over two and minus one can. This is from n equals one to infinity to buy the alternating signed test. This does give us convergence. It will include that. And then for X equals one, we have some from n equals one to infinity of one over two and minus one. And if we make the denominator something that's bigger, we should get something smaller. So this has to be bigger than one over to N. And this is half of the harmonic Siri's so half of infinity, which is still infinity. So this diverges case, we do not include one, but we do include minus one. So our interval of convergence is minus one and we are including that so close bracket and one give us divergence, so we're not including that.

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