Supremum of $A = \{ \frac{5^{3n}*5^{2m}}{3*5^{5n}+5^n + 2*5^{5m}} , n, m \in \mathbb{N}\}$.

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How do I find the supremum and infimum of set A?

$$A = \{ \frac{5^{3n}*5^{2m}}{3*5^{5n}+5^n + 2*5^{5m}} , n, m \in \mathbb{N}\}$$

I know that this expression is always greater than 0 since it is made of natural numbers only.

  • for $n = 1$, $m \implies \infty$, $A \implies 0$
  • for $m = 1$, $n = \implies \infty$, $A \implies 0$

Therefore, I know that the infimum is equal to $0$. However, how do I find the supremum?

From a graph I made I know that for $n = m = 1$ that expression is equal to $\frac{1}{5}$ and that might be the supremum. However, how do I deduct that from this expression? I tried dividing nominator and denominator by $5^n$ and $5^m$. I tried comparing nominator to denominator. Got nothing.

Any hint would be much appreciated.

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Hint : Let $x=5^n$ and $y=5^m$. Notice that

$$ \frac{1}{5}-\frac{x^3y^2}{3x^5+x+2y^5} = \frac{1}{5(3x^5+x+2y^5)}\bigg( x+(y-x)^2\big(3x^3 + 6yx^2 + 4y^2x + 2y^3\big)\bigg) $$