1.By integral test Since a(n)=1/[n*ln(n)*(ln lnn)^p] is decreasing and ...0 [ if n is sufficient large(in fact, n>3) and p>0], so we can use the integral test . ∫[a~∞] 1/[x ln(x) (ln lnx)^p] dx =∫[ln(lna)~∞] 1/u^p du ----(A) if p>...
Using integral test . Consider ∫ 1/x^5 dx (for 1 to ∞) =(-1/4)(1/x^4) | (1,∞) =-1/4 <∞ So, the original series converges.
一、use the Integral test 1. ∑(n=1~∞) 1/(n^2+1) Sol: 由∫(x=1...二、use the Comparison Test 1. ∑(n=1~∞) 1/n^2+n+1 Sol: 已知p級數...
...39;= lim(x->0) 14*49 / (6 + 36*49x^2) = 14*49 / (6 + 0)= 7*49/3= 343/3 (5-1) Integral Test ∫(1~oo)dx/x^2= -1/x= -1/oo + 1/1= 1= Convergence (5-2) Ratio...
最後一句是 Thus . by the integral test . ? 看起來未結束? 2008-05-08 14:56:50 補充： 如果p 小於或等於 0, 則級數中...
...否則就沒啥意思了 an=1/[nlnn*ln(lnn))],當n>=3時為正數,且遞減至0,可用 integral test 判別 積分結果為ln[ln(lnx)]代∞得∞,故divergent 4. n>=8時 lnn/&radic...
...series P級數 comparison test 比較審歛法 limit comparison test 極限比較審歛法 integral test 積分審歛法 absolutely convergent 絕對收歛 alternating series交錯級數 ratio test ...
