Let $a,b$ be in $\mathbb{R}$ such that $0 < a < b$. Prove that $\displaystyle \lim_{n \to \infty}(a^n + b^n)^{1/n} = b$
How would I go about proving this?
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1$\lim_{n \to \infty}$? $\lim_{n \to 0}$? Where does $n$ go to? – S.C.B. Mar 02 '17 at 09:02
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Where are you stuck ? – Mar 02 '17 at 09:03
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Hint: factor out $b^n$ – M. Winter Mar 02 '17 at 09:04
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@YvesDaoust I actually don't even know where to start. Any hints? – Adagio Mar 02 '17 at 09:04
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Very clearly, $b^n$ dominates $a^n$. How can you express this ? – Mar 02 '17 at 09:07
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Duplicate of (http://math.stackexchange.com/q/702498) itself a duplicate... – Jean Marie Mar 02 '17 at 09:45
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Hint: $(a^n + b^n)^{1/n} = b(1+\left(\frac{a}{b}\right)^n)^{1/n}$. And notice that $\frac{a}{b} < 1.$
rookie
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