93039

呢題點計係咪先假設P=1
之後計到個SERIES係DIVERGENT
再證當P大過1果陣係CONVERGENT就得

[ 本帖最後由 93039 於 2014-4-22 03:44 PM 編輯 ]

wsgoat

Hints onlya. basically need to prove p => q and q => p i.e. conv => p> 1 -- (1)  and p> 1 => conv  -- (2)
   (2) is simple
  for (1) can prove the contra+ve form i..e  (not p> 1) => not conv
  this mean proveing  p<=1 implies not conv
i.e. in addition to your proposal, you also need to prove div  for any p <1

b. use the fact that x(lnx)^p is increasing in the range k <= x <= k+1

93039

引用:

原帖由 wsgoat 於 2014-4-22 04:07 PM 發表
Hints onlya. basically need to prove p => q and q => p i.e. conv => p> 1 -- (1)  and p> 1 => conv  -- (2)
   (2) is simple
  for (1) can prove the contra+ve form i..e  (not p> 1) => not conv
  thi ...

如果我咁PROVE得唔得我計左個INTEGRAL出黎先
然後指當p=1,definite integral唔exist
然後就話div at p=1
但當P>1果陣,definite integral exist
所以conv. at p>1

cwk7093942

可以, 但你 in 錯

想知錯乜, 就自己對照公式

另外, 乜野叫 +ve constant ???

亦即係 p > 0

你淨係 prove p = 1 and p > 1

咁 0 < p < 1 呢, 去左邊......???????

桃子

Proving the case for x=1 does not answer the question.