- #1
Alcubierre
- 80
- 0
Homework Statement
The coefficients of the power series [itex]\sum_{n=0}^{∞}[/itex][itex]a_{n}(x-2)^{n}[/itex] satisfy [itex]a_{0} = 5[/itex] and [itex]a_{n} = (\frac{2n+1}{3n-1})a_{n-1}[/itex] for all [itex] n ≥ 1 [/itex]. The radius of convergence of the series is:
(a) 0
(b) [itex]\frac{2}{3}[/itex]
(c) [itex]\frac{3}{2}[/itex]
(d) 2
(e) infinite
Homework Equations
The Attempt at a Solution
[itex]lim_{n\rightarrow ∞}|\frac{a_{n+1}(x-2)^{n}}{a_{n}(x-2)^{n}}|[/itex]
[itex]|x-2|lim_{n\rightarrow ∞} |\frac{a_{n+1}}{a_{n}}|[/itex]
To find the function inside the limit, I used the definition presented in the problem:
[itex]a_{n} = (\frac{2n+1}{3n-1})a_{n}[/itex]
Added n + 1 and divided the [itex]a_{n}[/itex] from the RHS which yields,
[itex]\frac{a_{n+1}}{a_{n}} = \frac{2n+3}{3n+2} [/itex]
and substituted that into the limit, and took the limit:
[itex]\frac{2}{3}|x-2| < 1[/itex]
[itex]2|x-2| < 3[/itex]
[itex] 2[-(x-2)] < 1\; and\; 2(x-2) < 2 [/itex]
And that gives the interval,
[itex] x > \frac{1}{2}\; and\; x < \frac{7}{2} [/itex]
which means the radius is 2, but the correct answer is letter c, [itex]\frac{3}{2}[/itex].
What am I doing wrong? Was my approach in the right direction?
I solved this before and I used the rationale that the radius of convergence is the reciprocal of the limit of the interval, but my teacher said that wasn't correct and it was "one of those special cases that it works out that way."
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