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This question already has an answer here:

Let $a\geq 0$ and $ b\geq 0$. Prove that $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}$.
[Hint: Use the identity $(a^n -b^n)=(a-b)(\sum_{i=0}^{n-1}a^ib^{n-1-i})$]

I need some help! I cannot do it even with the hint... :(

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marked as duplicate by user147263, jameselmore, Siminore, Martin Sleziak, Jyrki Lahtonen Oct 3 '15 at 18:31

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    $\begingroup$ I think it's simpler to assume that $a>b$ and show that $\lim_{n\to\infty} a \sqrt[n]{1+(b/a)^n}=a$. $\endgroup$ – Myself Jan 19 '15 at 20:46
  • $\begingroup$ Do as @Myself says. $\endgroup$ – AlonAlon Jan 19 '15 at 20:47
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    $\begingroup$ Hehe, the comments read funny. "Do as Myself says. Thanks Myself. I'll try it." $\endgroup$ – Mark Fantini Jan 19 '15 at 20:51
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    $\begingroup$ Definitly ! Some schizophrenia in there :D $\endgroup$ – servabat Jan 19 '15 at 20:52
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Hint: $$\max\{a,b\} \leq \sqrt[n]{a^n+b^n} \leq \sqrt[n]{2\max\{a,b\}^n} = 2^{1/n}\max\{a,b\}$$ and make $n \to +\infty.$ Here's a bit more general version of this result.

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Assume without loss of generality $a > b$. Then you get

$$\sqrt[n]{a^n+b^n} = \sqrt[n]{a^n\cdot(1 + \frac{b^n}{a^n})} = a\cdot \sqrt[n]{(1 + \frac{b^n}{a^n})},$$

now conclude the limit using that $(\frac{b}{a})^n$ tends to $0$ for $n\to \infty$.

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Hint :suppose $a>b$, $n$ is odd $$a^{n}+b^{n}=(a+b)(a^{n-1}+a^{n-2}b^1+a^{n-3}b^2+...+b^{n-1})\\(a+b)(a^{n-1}+a^{n-2}a^1+a^{n-3}a^2+...+a^{n-1})\\\leq(a+a)(na^{n-1})=2na^n\\\sqrt[n]{a^{n}+b^{n}}\leq \sqrt[n]{2na^n}\rightarrow a $$

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