Question

第三，四题怎么做啊！哪位大神帮下忙，急！！！

Answer 1

(3)
(1- 1/2²)(1-1/3²)...(1- 1/n²)
=(1+ 1/2)(1- 1/2)(1+ 1/3)(1- 1/3)...(1+ 1/n)(1- 1/n)
= (3/2)(1/2)(4/3)(2/3)...[(n+1)/n][(n-1)/n]
=[3×4×...×(n+1)/(2×3×...×n)][(1×2×...×(n-1)/(2×3×...×n)]
=(n+1)/2](1/n)
=(n+1)/(2n) (之所以先化简，因为这个化简是初一的基础知识)
lim [(1- 1/2²)(1-1/3²)...(1- 1/n²)]
n→∞
=lim [(n+1)/(2n)]
n→∞
=lim [(1+ 1/n)/2]
n→∞
=(1+0)/2
=½
(4)
lim [(2ⁿ+3ⁿ)/(2ⁿ⁺¹+3ⁿ⁺¹)]
n→∞
=lim ⅓[(3·2ⁿ+3ⁿ⁺¹)/(2ⁿ⁺¹+3ⁿ⁺¹)]
n→∞
=lim ⅓[(2ⁿ⁺¹+3ⁿ⁺¹+2ⁿ)/(2ⁿ⁺¹+3ⁿ⁺¹)]
n→∞
=lim ⅓[1+ 2ⁿ/(2·2ⁿ+3·3ⁿ)]
n→∞
=lim ⅓[1+ 1/(2+3·(3/2)ⁿ)]
n→∞
=⅓(1+0)
=⅓