Solution:Using the definition we have,Let an=n−1limn→∞an=limn→∞n−1=Lthere exists an ϵ such that for all N>0,there is an n>N with ∣an−L∣≥ ϵ ................................................(1)Since when we put n=1,2,3,4,.......we get sequence {−1,0,1,2,3,4,......}when we put limn→∞ it approches to ∞.∴,by (1), sequencce {an} diverges.
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