If |x|<1 then limn→∞ (1+x)1+x21+x4…1+x2n is equal to
1x−1
11−x
1−x
x−1
We have,
limn→∞ (1+x)1+x21+x4…1+x2n=limn→∞ (1−x)(1+x)1+x2…1+x2n1−x.=limn→∞ 1−x4n1−x=1−01−x=11−x ∵limn→∞ xn=0 for −1<x<1