Let xnn1 be a sequence of real numbers such that xn does not

Let (x_n)_n=1 be a sequence of real numbers such that x_n does not equal 0 for all n N and x_n x as n , where x does not equal 0. Prove that inf ({|x_n| : n N}) > 0.

Solution

A sequence (x_n) of real numbers converges to a limit x R, written x = lim n xn, or x_n x as n , if for every > 0 there exists N N such that |x_n x| < for all n > N.

A sequence converges if it converges to some limit x R, otherwise it diverges.

Although we don’t show it explicitly in the denition, N is allowed to depend on .

Typically, the smaller we choose , the larger we have to make N.

One way to view a proof of convergence is :

If > 0, you have to come up with an N that “works.” Also note that x_n x as n means the same thing as

|x_n x| 0 as n . It may appear obvious that a limit is unique if one exists, but this fact requires proof.

proof 2:

Suppose that (xn) is a sequence such that xn x and x_n x0 as n . Let > 0.

Then there exist N,N0 N such that |x_n x| < 2 for all n > N, |xn x0| < 2 for all n > N0.

Choose any n > max{N,N0}. Then, by the triangle inequality, |xx0| |xxn| + |xn x0| < /2 + / 2 < .

Since this inequality holds for all > 0, we must have |xx0| = 0 (otherwise the inequality would be false for = |xx0|/2 > 0),

so x = x0. The following notation for sequences that “diverge to innity” is convenient.