let an=(-1)^(2n+1) and L=-1 let us ahow that L=lim an indeed if e >0 we must find N such that if n>=N then |an+1|< e that is examine the quantity |an+1| find the natural number N |an+1|=|(-1)2n+1+1| =|(-1)2n+(-1)1+1| =|1-1+1|=1=n thus |an+1| <= 1/n therefore n is an integer for which N>1/e therefore for every e>0 there exist a natural number N such that n>=N then |an-L|<1/n<1/N infinity [(-1)2n+1] = -1 Solution let an=(-1)^(2n+1) and L=-1 let us ahow that L=lim an indeed if e >0 we must find N such that if n>=N then |an+1|< e that is examine the quantity |an+1| find the natural number N |an+1|=|(-1)2n+1+1| =|(-1)2n+(-1)1+1| =|1-1+1|=1=n thus |an+1| <= 1/n therefore n is an integer for which N>1/e therefore for every e>0 there exist a natural number N such that n>=N then |an-L|<1/n<1/N infinity [(-1)2n+1] = -1.