Re: [微積] 一題極限
※ 引述《cxcxvv (delta)》之銘言:
: Show that
: if X_n - X_(n-2) → 0 (n→∞)
: then X_n/n → 0
X = (X - X ) + (X - X ) + ... + (X - X ) + X
n n n-2 n-2 n-4 k+2 k k
where k = 1 if n is odd and k = 2 if n is even. Since
X - X → 0 as n→∞, for any ε > 0, we can find N
n n-2
ε
such that │X - X │ < ── for all n ≧ N. Next, we can find
n n-2 2
│X_k+( X_(k+2)-X_k )+...+( X_N-X_(n-2) )│
natural number M such that │──────────────────────│
│ n │
ε
< ── for all n ≧ M > N.
2
│ X_n │ │ X_k + ( X_(k+2) - X_k ) + ... + ( X_N -X_(N-2) )│
So │───│ < │───────────────────────────│
│ n │ │ n │
[n - N] ε ε
+ ───── ε < ── + ── = ε for all n ≧ M.
2n 2 2
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09/18 21:48, , 1F
09/18 21:48, 1F
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