show that if $\displaystyle 0<a<b, then (a^n + b^n)^{1/n} --> b$
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Originally Posted by wopashui show that if $\displaystyle 0<a<b, then (a^n + b^n)^{1/n} --> b$ $\displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$
Originally Posted by Plato $\displaystyle b = \sqrt[n]{{b^n }} \leqslant \sqrt[n]{{b^n + a^n }} \leqslant \sqrt[n]{{b^n + b^n }} = b\sqrt[n]{2}$ this just proved it's bounded, is it?
Originally Posted by wopashui this just proved it's bounded, is it? What is $\displaystyle \lim_{n\to\infty} b$ ? What is $\displaystyle \lim_{n\to\infty} b \sqrt[n]{2}$ ? What does that tell you then if $\displaystyle b \le \sqrt[n]{a^n + b^n} \le b\sqrt[n]{2}$ ?
Originally Posted by wopashui this just proved it's bounded, is it? Apply the squeeze theorem.
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