Friday, October 4, 2019

real analysis - Show that limnrightarrowinftysqrt[n]cn1+cn2+ldots+cnm=maxc1,c2,ldots,cm





Let mN and c1,c2,,cmR+. Show that lim



My attempt: Since \lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} \leq \lim_{n\rightarrow \infty}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}} = \lim_{n\rightarrow \infty} \sqrt[n]{n}\sqrt[n]{\max\{c_1,c_2,\ldots,c_m\}}=\lim_{n \rightarrow \infty} \max\{\sqrt[n]{c_1^n},\sqrt[n]{c_2^n},\ldots,\sqrt[n]{c_m^n}\}=\lim_{n \rightarrow \infty}\max\{c_1,c_2,\ldots,c_m\}=\max\{c_1,c_2,\ldots,c_m\}



it follows that \lim_{n\rightarrow \infty} \sqrt[n]{c_1^n+c_2^n+\ldots+c_m^n} is bounded, but I don't think it's monotonically decreasing, at least I can't prove this. Can anybody tell me whether the approach I have chosen is a good one, whether what I have done is correct and how to finish the proof?



Answer



The short proof.



Let c=\max\{c_1,c_2,\dots,c_m\} and note that:



c^n \leq c_1^n+c_2^n+\dots+c_m^n \leq mc^n



Now take the nth root, and see that \lim_{n\to\infty} \sqrt[n]{m} = 1.


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