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danni7070
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[Solved] How to calculate.
Determine the centre, radius, and Interval of convergence of the power series.
[tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]
Radius of convergence
[tex] \frac{1}{R} = L = \lim_{n \to {\infty}} \frac{a_n+1}{a_n} [/tex]
so R = 1/L
and the interval is (x-R,x+R)
Finding the centre is easy. x = -2
But the algebra stuff is getting me down. This is where I get confused. I took this from the solution manual and I can't figure out this result.
[tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]
So we have [tex] R = \lim \frac{2^{n+1}(n+1)}{2^n n} = 2 [/tex]
Homework Statement
Determine the centre, radius, and Interval of convergence of the power series.
[tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]
Homework Equations
Radius of convergence
[tex] \frac{1}{R} = L = \lim_{n \to {\infty}} \frac{a_n+1}{a_n} [/tex]
so R = 1/L
and the interval is (x-R,x+R)
The Attempt at a Solution
Finding the centre is easy. x = -2
But the algebra stuff is getting me down. This is where I get confused. I took this from the solution manual and I can't figure out this result.
[tex] \sum_{n=1}^\infty \frac{1}{n} \frac{(x+2)}{2}^n [/tex]
So we have [tex] R = \lim \frac{2^{n+1}(n+1)}{2^n n} = 2 [/tex]
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