Prove thisIf {sn} , n=1 to infinity is a sequence of real numbers and if lim(n tends to infinity)sn = M , where M is not equal to , then lim(n tends to infinity) 1/sn = 1/M

Answer 1

The correct formulation of this problem is: if a sequence (s_n) has a non-zero limit M, then the sequence of reciprocals (s_n^-1) has limit M^-1. Here is a proof by the so-called "ε-N" argument.

By definition, (s_n) has limit M means ∀ε>0, ∃N∈N (N denotes the set of natural numbers) such that ∀n>N, |s_n − M|<ε.

Our goal is to prove that ∀ε'>0, ∃N'∈N, such that for any n>N', |s_n^-1−M^-1|<ε' (or a multiple of ε, say kε, where k is a finite positive number independent of n or ε). The key lies in the fact that M≠ 0. Without loss of generality, we may assume that M>0. Now for our given ε'>0, to ensure that |s_n^-1−M^-1|<ε', we try to solve for n: |s_n^-1−M^-1|<ε' is equivalent to |s_n − M|/|s_nM|<ε'. It suffices to show that ε/|s_nM|<ε' or ε<|s_nM|ε'. The question now is how large |s_nM| can be. If we choose ε₁=M/2, then for this particular ε₁, there is an N₁∈N, such that for any n>N_1, |s_n − M|<ε₁=M/2. By a version of the triangle inequality, ||s_n|− |M|| ≤ |s_n − M| < M/2. Thus we can at least conclude that |s_n| > M − M/2=M/2. it follows that |s_n − M|/|s_nM|<ε/(M²/2). Notice that |s_n − M|<ε in the numerator is achieved when n>N, whereas |s_nM|< M²/2 in the denominator is achieved when n>N_1. So to achieve both, we need to have n>max(N, N₁). Now we still can choose our ε when ε' is given. Since we want to have

|s_n^-1−M^-1| = |s_n − M|/|s_nM| < ε/(M²/2) can convert to |s_n^-1−M^-1| < ε', it suffices to make ε/(M²/2) = ε', or ε = ε'(M²/2).

To summarize, for any ε'>0, let ε = ε'(M²/2), and ε₁=M/2. Then there exists an N and an N₁, for any n>N, |s_n − M|< ε = ε'(M²/2), while for any n>N₁, |s_n| > M/2. Now let N'=max(N, N₁). Then for any n>N', we have

|s_n^-1−M^-1| = |s_n − M|/|s_nM| < ε/(M²/2)=ε'(M²/2)/(M²/2) = ε'. By definition, (s_n^-1) has limit M^{-1, as desired.}