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

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?