※ 引述《plover (>//////<)》之銘言：
: ∞ 1
: Consider the series Σ ( ----- - √( ln(1 + 1/n) ) ).
: n=1 √n
: Is it convergent?
原級數收斂 !! 證明之前先給一個事實 ln(1+1/x) < 1/x for any x≧1
故 (1/√n) - √( ln(1 + 1/n) ) > (1/√n) - (1/√n) = 0 for nature number n
因此原級數是正項級數 ... 用極限比較測試與 1/n^(3/2) 作比較
lim {[1-√(nln(1 + 1/n))]/n^(1/2)} / [1/n^(3/2)]
n->oo
= lim {[1-√(nln(1 + 1/n))] / n^(-1)
n->oo
= lim {(1/2)[nln(1 + 1/n)]^(-1/2)} {[ln(1 + 1/n)] - 1/(n+1)} / [n^(-2)]
n->oo
上色部分 -> 1/2 當 n -> oo, 故僅需考慮
lim {[ln(1 + 1/n)] - 1/(n+1)} / [n^(-2)]
n->oo
= lim [-1/n(n+1) + 1/(n+1)^2] / (-2) n^(-3)
n->oo
= (1/2) lim [1/n(n+1)^2] / n^(-3)
n->oo
= (1/2) lim [n/(n+1)]^2 = 1/2
n->oo
故原極限值 = 1/4 且 Σ 1/n^(3/2) 收斂
故原級數收斂
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