Find the supremum and infimum of this set

Question

Find the l.u.b. and g.l.b. of

{$2^{-p}$ + $3^{-q}$ + $5^{-r}$ $\in$ $\Bbb Q$ : $p, q, r$ $\in$ $\Bbb Z^+$}

I believe I correctly found supE = $^{31}$/$_{30}$

Since $p, q, r$ $\in$ $\Bbb Z^+$, we have $2^{-p}$, $3^{-q}$, $5^{-r}$ at their greatest values would be $^{1}$/$_{2}$, $^{1}$/$_{3}$, $^{1}$/$_{5}$ since 1 is the smallest element in $\Bbb Z^+$.

So, $2^{-1}$ + $3^{-1}$ + $5^{-1}$ = $^{31}$/$_{30}$

However, I have no idea how the infE = 0 ... can somebody please explain?

Answer 1

What you did is correct. And $\inf E=0$. It is claer that each element of $E$ is greater than $0$ and, on the other hand$$\lim_{n\to\infty}2^{-n}+3^{-n}+5^{-n}=0.$$

Answer 2

Clearly the elements are positive.

Also, notice that $\lim_{p \to \infty} 2^{-p} = 0$, Similar result for the other two terms.

Hence we can make the terms arbitrarily small but positive. Hence the infimum is $0$.