※ 引述《master791217 ( )》之銘言：
: find the radius of convergence of power series
: ∞ １ 2k
: 1 Σ 一一 （ｘ＋１）
: k=0 ４^k ans: R=2 #
: ∞ 1 k^2 k
: 2 Σ （１ － 一一 ） Ｘ ans: R= e #
: k=1 k
: 懇求高微高手
: 拜託幫忙解題
: 需要完整的解題的過程
: 非常感謝~!
2k
Sol: (x+1) a 2(k+1) k+1
1. Let a = --------- => k+1 (x+1) / 4
k k ------- = -------------------
4 a 2k k
k (x+1) / 4
(2k+2-2k) (k+1-k)
= (x+1) / 4
2
= (x+1) / 4
=> By ratio test,
2 2
lim | a / a | = lim |(x+1) / 4| = |(x+1) / 4|
k->∞ k+1 k k->∞
2
若 | (x+1) / 4 | < 1 ,則必收斂
2
=> |x+1| < 4
=> |x+1| < 2
若|x+1| = 2 => x = -3 or 1時,都不會收斂,可直接代進去驗證
2k k
x = -3 => a = (-2) / 4 = 1 ∞
k 2k k => Σ 1 -> ∞
x = 1 => a = 2 / 4 = 1 k=1
k
所以收斂半徑 = 2
2 2
2. 1 k k 1/k 1 k k 1/k
Let b = (1- ---) x => b =[ (1- ---) x ]
k k k k
2
1 k 1/k k 1/k
= [(1- ---) ] ( x )
k
2
k *1/k k*1/k
= (1-1/k) x
k
= (1-1/k) * x
=> by root test,
k k
lim |(1-1/k) * x | = lim |(1-1/k)| * lim |x|
k->∞ k->∞ k->∞
-u
= | lim (1+1/u) | *|x| (Let u = -k)
k->∞
u -1
= |[lim (1+1/u) ] | *|x|
-1
= e |x|
-1
若 e |x| < 1 => |x| < e 時必收斂
所以收斂半徑 = e
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