assume that lim[1+2(-1)^n]Xn = 0. Prove that lim Xn exists and find it.
1
Expert's answer
2012-10-16T09:48:39-0400
if n=2k then lim[1+2(-1)^n]Xn=lim[1+2(-1)^(2k)]X2k=lim3*X_(2k) if n=2k+1 then lim[1+2(-1)^(2k+1)]Xn=lim(-1)*X_(2k+1) then for subsequences 0=lim3*X_(2k)=3lim X_(2k) and 0=lim(-1)*X_(2k+1)=-lim X_(2k+1)
so we have lim X_(2k)=lim X_(2k+1)=0 so limit exist and equal 0
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