推 math1209:p > 1. 用比較審斂法就行了. 114.32.219.116 04/12 19:43
→ midarmyman:怎麼比? 140.117.198.78 04/12 20:20
→ math1209:我只解釋 p > 1, 至於為什麼 p <= 1 不 114.32.219.116 04/12 20:23
→ math1209:會使其收斂, 你自己想想看 XD 114.32.219.116 04/12 20:24
→ math1209:當 p > 1 寫 p = 1 + a, a > 0. 114.32.219.116 04/12 20:24
→ math1209:於是 (ln n)/n^p = 114.32.219.116 04/12 20:25
→ math1209:{(ln n)/n^(a/2)}*{1/n^(1 + a/2)}. 114.32.219.116 04/12 20:25
→ math1209:顯然, {(ln n)/n^(a/2)} -> 0 as n ->oo. 114.32.219.116 04/12 20:26
→ math1209:因此當 n 夠大時, {(ln n)/n^(a/2)} < 1. 114.32.219.116 04/12 20:27
→ math1209:於是所求級數之斂散可看成 Σ1/n^(1+a/2) 114.32.219.116 04/12 20:27
→ math1209:之斂散. 這顯然收斂 XD. 114.32.219.116 04/12 20:27
→ math1209:積分測試法也是可行的 XD... 114.32.219.116 04/12 20:29
→ midarmyman:謝了! 好方法! 140.117.198.78 04/12 20:31