Re: [考古] 淡江96
※ 引述interval of convergence《joyce123 radius of convergence(星稀)》之銘言：
: Find the radius of convergence(收歛半徑) and interval of convergence (收歛區間
: n n
: ∞ (-5) x
: ) of the series Σ -----------
: n=0 √(n+1)
: 請各位高手解答~ 感謝^^
Let a_n = (-5)^n x^n / √(n+1)
a_n+1 √(n+1)
---- = -5x --------
a_n √(n+2)
lim |a_n+1/a_n| = |5x|
n->oo
By ratio test , |5x| < 1 converges.
=> |x|< 1/5 , the radius of convergence is 1/5
If x=1/5 ,
a_n = (-1)^n/√(n+1)
By 交錯級數test, we know that Σa_n converges
If x=-1/5
_____
=> a_n = 1/√(n+1)
√(n+1) < √(n+n)
1/√(n+1) > 1/(√2*√n)
Σ1/√(n+1) > Σ1/(√2*√n)
By p-series test , we know that Σ1/(√2*√n) diverges.
By comparison test , we know that Σ1/√(n+1) diverges.
Hence , the interval of convergence is (-1/5 , 1/5]
有錯請指正
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