Re: [張爸] 級數1
※ 引述《GLP (^__________^)》之銘言:
: 求級數和
: ∞ n^2 - n - 1
: Σ ------------
: n=1 n!
∞ n^2 ∞ n ∞ k + 1
Σ --------- = Σ -------- = Σ -------- = 2e
n=1 n! n=1 (n-1)! k=0 k!
∞ n ∞ 1
Σ ------ = Σ -------- = e
n=1 n! n=1 (n-1)!
∞ 1
Σ ------- = e-1
n=1 n!
∞ n^2 - n - 1
合併上面 得到 Σ --------------- = 2e - e - e + 1 = 1
n=1 n!
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