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the radius of convergence for the power series [itex]\sum[/itex](-1)^n [itex]\frac{(x-3)^n}{(n+1)}[/itex] is R=1. Determine the interval of convergence.

What does this mean? can anyone help me solve this please

- Thread starter xtrubambinoxpr
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- #1

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the radius of convergence for the power series [itex]\sum[/itex](-1)^n [itex]\frac{(x-3)^n}{(n+1)}[/itex] is R=1. Determine the interval of convergence.

What does this mean? can anyone help me solve this please

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HallsofIvy

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To determine for what values of x the

"ratio test"- The numerical sum of

Of course, the numbers, for all x, are not necessarily positive so we have to talk about "absolute convergence", the convergence of [itex]\sum |a_n x^n|[/itex].

For example, the ratio of of the form [itex]|a_{n+1}(x-x_0)^{n+1}|/|a_n x^n|[/itex] reduces to [itex](a_{n+1}/a_n)|x- x_0|[/itex]. Saying the limit is less than 1 is exactly the same as saying that [itex]|x-x_0|< \lim_{n\to\infty}|a_n/a_{n+1}[/itex].

That is, if we let [itex]r= \lim_{n\to\infty}|a_n/a_{n+1}|[/itex], then we are saying that the series converges for [itex]|x-x_0|< r[/itex] which is, of course, the same as saying [itex]-r< x- x_0< r[/itex] or [itex]x_0- r< x< x_0+ r[/itex]. That is precisely saying that the series converges for x in the open interval [itex](x_0- r, x_0+ r)[/itex] and diverges outside the interval [itex]x_0- r\le x\le x_0+ r[/itex] (the endpoints, [itex]x_0- r[/itex] and [itex]x_0+ r[/itex], have to be checked separately). The interval [itex](x_0- r, x_0+ r)[/itex], including one or both endpoints if the series converges there, is called the "interval of convergence" and "r" is called the "radius of convergence"

(When you work with power series or

Now, if we use the ratio test to determine the convergence of [itex]\sum (-1)^n (x-3)^n/(n+1)[/itex], we look at the ratio

[tex]\dfrac{|x- 3|^{n+1}}{n+2}\dfrac{n+3}{|x-3|^n}= \frac{n+2}{n+3}|x-3|[/tex]

You should ge able to see that the limit, as n goes to infinity, of (n+2)/(n+3) is 1 so we know that this series converges absolutely for |x- 3|< 1. That is the same as saying that -1< x- 3< 1 or 3-1= 2< x< 3+ 1= 4. It diverges outside the interval. Whether or not it converges at x= 2 or x= 4 would have to be decided separately.

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