※ 引述《stacyfish (欣)》之銘言：
: n
: lim 1/(n^k+1) Σ i^k = 1/(k+1)
: n→∞ i=1
: n
: 該題給的hint是 Σ [i^(k+1)-(i-1)^(k+1)] = n^(k+1)
: i=1
: 有嘗試使用數學歸納法~
: 但還是找不到hint該使用在什麼地方?
: 想請板上大大解惑 謝謝!
n
Let H(k) = Σ i^k , H(0) = n
i=1
n
Consider (n+1)^(k+1) -1 = Σ (i+1)^(k+1) - i^(k+1)
i=1
n k+1 k+1
= Σ [ Σ C i^(k+1-j) - i^(k+1) ] (binomial)
i=1 j=0 j
n k+1 k+1
= Σ Σ C i^(k+1-j) (j=0的term與i^(k+1)相消)
i=1 j=1 j
k+1 n k+1
= Σ Σ C i^(k+1-j)
j=1 i=1 j
k+1 k+1 n
= Σ C Σ i^(k+1-j)
j=1 j i=1
k+1 k+1
= Σ C H(k+1-j)
j=1 j
k+1 k+1
= (k+1)H(k) + Σ C H(k+1-j)
j=2 j
k+1 k+1
(n+1)^(k+1) - 1 - Σ C H(k+1-j)
j=2 j
so H(k) = ────────────────── ----(*)
k+1
we find deg H(1) = 2 since deg H(0) = 1
H(2) = 3 since def H(1) = 2
by induction , we have deg H(k) = k+1
Again from (*) , since deg H(k+1-j) = k+2-j , j=2 to k+1
(n+1)^(k+1) + a_k*n^k + a_(k-1)*n^(k-1) + ... + a_0
H(k) = ───────────────────────────
k+1
H(k) (n+1)^(k+1) + a_k*n^k + a_(k-1)*n^(k-1) + ... + a_0
so, ──── = ───────────────────────────
n^(k+1) n^(k+1) * (k+1)
1 1
= ─── * (1+ ──)^(k+1)
k+1 n
1 1 1 1
+ ─── * (a_k*── + a_(k-1)*─── + ... + a_0*────)
k+1 n n^2 n^(k+1)
take n → +inf , done
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