How to show $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}$?

Question

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... :(

Answer 1

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.

Answer 2

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 $n\to \infty$.

Answer 3

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