# \$limsup{1over x_n}={1over liminf x_n}\$?

Prove that if there exists a constant \$a\$ such that \$0<ale x_n\$ then \$\$limsup{1over x_n}={1over liminf x_n}\$\$. \${x_n}\$ is bounded.

It is important for me to know where exactly I am wrong. Here is my attempt:

\$x_n\$ is lower bounded and therefore must have a minimal partial limit, \$m=lim inf x_n\$ (I don’t know how to explain why formally)

\$x_n\$ is upper bound then it must have a maximal partial limit (I don’t know how to explain why formally), \$s=lim sup x_n\$. Then \$limsup {1over x_n}=max PL{{1over m},…,{1over s}}={1over m}={1over min PL{m,…,s}}={1over liminf x_n}\$.

Why do I actually need \$a\$? I would appreciate your reply.

## Answer

Since \$x_n ge a > 0\$ for all \$n\$, \$liminf x_n ge a > 0\$, hence \$1/liminf x_n\$ is a finite number. Now since \$frac{1}{x_n} le frac{1}{a}\$, \$limsup frac{1}{x_n} le frac{1}{a} < infty\$ So the equation above at least makes sense. Without having that \$a\$ present in the hypothesis, you can have a \$1/0\$ situtation (e.g., if \$x_n = 1/2^n\$), which is problematic.

To show that the equation holds, let \$L = limsup 1/x_n\$ and \$M = liminf x_n\$. Given \$epsilon > 0\$, \$1/x_n < L + epsilon\$ for all but finitely many \$n\$. Thus

\$\$x_n > frac{1}{L + epsilon}\$\$

for all but finitely many \$n\$. Hence

\$\$liminf x_n ge frac{1}{L + epsilon},\$\$

that is, \$M ge 1/(L + epsilon)\$. On the other hand, by definition of \$M\$, \$x_n > M – epsilon\$ for all but finitely many \$n\$. Thus

\$\$frac{1}{x_n} < frac{1}{M – epsilon}\$\$

for all but finitely many \$n\$. Therefore

\$\$limsup frac{1}{x_n} le frac{1}{M – epsilon},\$\$

or \$L le frac{1}{M – epsilon}\$. Letting \$epsilon to 0^+\$ results in \$M ge 1/L\$ and \$L le 1/M\$. In other words, \$L ge 1/M\$ and \$L le 1/M\$. Therefore \$L = 1/M\$, as desired.