1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)
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1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)
![1/(2!)+1/(3!)+1/(4!)+1/(5!) +…+1/(n!)](/uploads/image/z/18173087-71-7.jpg?t=1%2F%282%21%29%2B1%2F%283%21%29%2B1%2F%284%21%29%2B1%2F%285%21%29+%2B%E2%80%A6%2B1%2F%28n%21%29)
放缩法
n>3时,n!=1*2*...*(n-1)*n>(n-1)n
所以原式
n>3时,n!=1*2*...*(n-1)*n>(n-1)n
所以原式
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
用数学归纳法证明:1×2×3+2×3×4+…+n×(n+1)×(n+2)=n(n+1)(n+2)(n+3)4(n∈N
lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2)
(1/(n^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大
lim 9^n+4^n+2/5^n-3^2n-1 n趋于无穷大时
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?
化简:1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
(n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1
证明(1+2/n)^n>5-2/n(n属于N+,n>=3)
一道数列求和题1/2n+3/4n+5/8n+...+(2n-1)/n*2^n
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[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简