Respuesta :
We can find the power series representation of the Taylor series, and the radius and interval of convergence.
Taylor series
Since we already have the chart done, the value in the far right column becomes the coefficient on each term in the Taylor polynomial, in the form
\frac{f^{(n)}(a)}{n!}(x-a)^nn!f(n)(a)(x−a)n
To find the interval of convergence, we’ll take the inequality we used to find the radius of convergence, and solve it for xx.
|x-3|<3∣x−3∣<3
-3<x-3<3−3<x−3<3
-3+3<x-3+3<3+3−3+3<x−3+3<3+3
0<x<60<x<6
We need to test the endpoints of the inequality by plugging them into the power series representation. We’ll start with x=0x=0.
The exponent on the -1−1 will always be odd, so the sum is going to simplify to
\sum^{\infty}_{n=1}-\frac{1}{n}∑n=1∞−n1
This is a divergent pp-series, so the series diverges at the endpoint x=0x=0. Now we’ll test x=6x=6.
\sum^{\infty}_{n=1}\frac{(-1)^{n+1}(6-3)^n}{n3^n}∑n=1∞n3n(−1)n+1(6−3)n
\sum^{\infty}_{n=1}\frac{(-1)^{n+1}3^n}{n3^n}∑n=1∞n3n(−1)n+13n
\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}∑n=1∞n(−1)n+1
This is an alternating series where
a_n=\frac{1}{n}an=n1
The alternating series test for convergence says that a series converges if \lim_{n\to\infty}a_n=0limn→∞an=0.
\lim_{n\to\infty}\frac{1}{n}limn→∞n1
\frac{1}{\infty}∞1
00
The series converges at the endpoint x=6x=6.
This shows that the series diverges at x=0x=0 and converges at x=6x=6, which means the interval of convergence is
0<x\le60<x≤6
To learn more about Taylor series refer: https://brainly.com/question/28158012
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