If available, you will find $ sup S, inf S $

$$ S: = Big { left (m + n right) ^ { frac {1} {mn}}: ; m, n in mathbb N Big } $$

$$ m = 1, n = 1: ; left (m + n right) ^ { frac {1} {mn}} = 2 $$

For a firm $ n in mathbb N, m in infty ; $:

$ displaystyle lim_ {m bis infty} (m + n) ^ { frac {1} {mn}} = lim_ {m bis infty} Bigm (m left (1+ frac { n} {m} right) Big) ^ { frac {1} {m} cdot frac {1} {n}} = lim_ {m to infty} left ( sqrt (m) {m} left (1+ frac {n} {m} right) ^ { frac {1} {m} right) ^ { frac {1} {n}} $$$= lim_ {m bis infty} sqrt (n) { sqrt (m) {m}} lim_ {m bis infty} left (1+ frac {n} {m} right) ^ { frac {1} {mn}} = e $$

$$ f (m): = left (1+ frac {1} {mn} right) ^ { frac {1} {mn}} $$

$$ m & # 39;> m implies f (m & # 39;)> f (m) $$

$$ m + n geq 2 (mn) ^ { frac {1} {2}} Big / ^ { frac {1} {mn}} $$

$$ (m + n) ^ { frac {1} {mn}} geq sqrt { sqrt (mn) {mn}} ; sqrt (mn) {2}> 1 $$

Then I concluded: $$ f (x) in langle 1, e rangle implies inf S = 1, sup S = e $$

Is that correct?