Σ ∞ n=1 ( n(n+1)^4)^-1

Σ ∞ n=1 ( n(n+1)^4)^-1 ∞ Σ 3n^n/(1+n)^n 判定收敛性 n=1∞ -------------------------------------------------------------------------------------------------------------Σ 3n^n/(1+n)^n ---------------------------------------------------------------------------------- ∞ Σ 3n^n/(1+n)^n 判定收敛性 n=1, (n+1)^n-(n-1)^n=? 推导 n*n!=(n+1)!-n! ∑(∞,n=1)(n^2-2n+3)/(n^4+n^2-6) 判断敛散性：∑(n=1~∞) (4^n*n!*n!)/(2n)! Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗? ∞Σ n^(-2)n=1 判断级数敛散性∑(n=1到∞)(n+1/n)/(n+1/n)^n 大一微积分解答：lim(n→+∞)（2^n+4^n+6^n+8^n）^1/n=? n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10这道题怎么解 n(n-1)(n-2)（n-3)（n-4)=154440.求N值 要步骤 求和Σ1/n^4,n=1->∞, 4^n+4^(n-1)C1/n+4^(n-2)C2/n+...+Cn/n=答案是5^n. 证明：1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+) 求极限 lim(n→∞) tan^n (π/4 + 2/n) lim(n→∞)tan^n(π/4+2/n) =lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n =lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n =lim(n→∞)(1+tan(2/n))^n/(1-tan(2/n))^n (1) 因为 lim(n→∞)(1+tan(2/n) lim(n->∞)[2^(2n-1)+1]/(4^n-3^n)=?