推 bineapple :推 02/09 22:53
推 WINDHEAD :戀曲 LA 02/10 01:24
推 tasukuchiyan:感謝 不過，第一個方法不懂 02/10 12:32
推 bineapple :第一個方法把(3k-2)/3k寫成1-2/3k 再取log 02/10 13:31
→ tasukuchiyan:log(1-2/3k)=Σ-((2/3k)^n)/n 這樣嗎? 再來是... 02/10 18:53
※ 引述《tasukuchiyan (Tasuku)》之銘言：
: 1.Σ(1．4．．．(3n-2)/(3．6．．．3n))^2
: 2.Σ(1．3．．．(2n-1)/(2．4．．．2n))^2
: 請問這兩題要怎麼驗證它是否收斂？有任何想法嗎？謝謝。
By using taylor expansion of log(1+x), one can see easily that
(1．4．．．(3n-2)/(3．6．．．3n))^2 = O(1/n^(4/3))
and
(1．3．．．(2n-1)/(2．4．．．2n))^2 = O(1/n).
So, the first series converges and the 2nd diverges.
For basic approach:
1 4 (3n-2)
1.let a= -- x --- x ....x--------
3 6 3n
2 5 (3n-1)
< b= -- x --- x ....x--------
4 7 3n+1
3 6 3n
< c= -- x --- x ....x-----.
5 8 3n+2
hence abc= 2/(3n+1)(3n+2),
so a^2< (abc)^(2/3) = O(1/n^(4/3)).
Thus, series in Q1 converges.
3*5*.......*(2n-1) 4*6*........*(2n)
2.let a= ----------------------- > b= -----------------------.
2*4*.......*(2n-2)*2n 3*5*........*(2n-1)*2n
Hence a^2> ab=1/4n.
Thus, series 2 diverges.
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