How do you prove if lim_(n->infty) x_(2n) = L = lim_(n->infty) x_(2n+1), then lim_(n->infty) x_n = L?

1 Answer
Jul 27, 2017

If lim_(n->oo) x_(2n) = L, this means that for any number epsilon_e > 0 we can find N_e such that:

n > N_e => abs(x_(2n)-L) < epsilon_e

On the other hand, lim_(n->oo) x_(2n+1) = L means that for any number epsilon_o > 0 we can find N_o such that:

n > N_o => abs(x_(2n+1)-L) < epsilon_o

Choose now: epsilon = min(epsilon_e, epsilon_o) and N = max(N_e,N_o)

Clearly for any n > N we have that if n is even, as n > N >= N_e:

abs(x_(n)-L) < epsilon_e <= epsilon

and if n is odd, as n > N >= N_o:

abs(x_(n)-L) < epsilon_o <= epsilon

Either way:

n > N => abs(x_(n)-L) < epsilon

which proves that:

lim_(n->oo) x_n = L