Re: 審歛

[znmkhxrw (QQ)]

※ 引述《min102257 (暱稱)》之銘言：
:      ∞         1
:      Σ   ---------------  conv.?
:      N=3        ln(lnN)
:             (lnN)
: 有點複雜不知怎麼算
: 化成積分好像也積不出來
: 那應該是用 比較檢驗法??
: 麻煩高手了~ 感謝

divergent.


since ln(ln(n)) is increasing , ln(n) is increasing

so
        ln(ln(n))
   ln(n)           is increasing to inf

         1
so  ─────── decreasing to 0
         ln(ln(n))
    ln(n)


Cauchy condensation principle can be applied

             2^n
consider   ─────────  (n都用2^n帶入  然後整個級數乘2^n)
                  ln(nln(2))
            nln(2)

             2^n
=>         ─────────────
                  (ln(n) + ln(ln(2)))
            nln(2)

             2^n
=>         ────────────────────────── denoted by b_n
             ln(n)    ln(ln(2))        ln(n)          ln(ln(2))
            n      * n          * ln(2)       *  ln(2)
                                  ~~~~~~~~~~
                                      ↓
                                     ln(ln(2))
                                    n


by root test, (b_n)^(1/n)

                 2
           =  ────────────────────────────────
               (ln(n))/n    (ln(ln(2)))/n    (ln(ln(2)))/n        (ln(ln(2)))/n
              n          * n              * n              * ln(2)
              ~~~~~~~~~~   ~~~~~~~~~~~~~~   ~~~~~~~~~~~~~~   ~~~~~~~~~~~~~~~~~
                 ↓              ↓               ↓                 ↓
as n→inf,        1              1                1                   1

so summation of b_n diverges

imply the original question is divergent

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