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I was thinking about the following problem:

How can i find the value of $(2^n+3^n+4^n)^{1/n}$ as $n \rightarrow \infty?$

Can someone point me in the right direction? Thanks in advance for your time.

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This is known as the infinity norm of the vector $(2, 3, 4)$. –  Joe Z. Jan 8 '13 at 13:43
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3 Answers

up vote 14 down vote accepted

HINT: $$4=(4^n)^{\frac1n}\leq (2^n+3^n+4^n)^{\frac1n}\leq (3\cdot4^n)^{\frac1n}=3^{\frac1n}\cdot4.$$


In general $$\displaystyle{\lim_{p\to\infty}(|x_1|^p+|x_2|^p+\cdots+|x_k|^p)^{\frac1p}=\max_{i=1,2,\ldots k}|x_i|}\cdot$$ This is the sup-norm of $(x_1,x_2,\cdots,x_k)$.
The notation is $\|(x_1,x_2,\cdots,x_k)\|_\infty$ because of the $$\displaystyle{\max_{i=1,2,\ldots k}|x_i|=\lim_{p\to\infty}(|x_1|^p+|x_2|^p+\cdots+|x_k|^p)^{\frac1p}=\lim_{p\to\infty}\|(x_1,x_2,\cdots,x_k)\|_p}\cdot$$

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You can take a factor of four out to give:$$4\times((\frac2 4)^n+(\frac3 4)^n+1)^{\frac1n}$$ and work from there.

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Thanks a lot sir.Got it.. –  learner Jan 8 '13 at 13:49
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Here you have an approach that uses the definition of the constant $e=\lim_{n\to\infty}(1+\frac 1n)^n$. $$ (2^n+3^n+4^n)^{\frac 1n} ~ = ~~ 4 \left( 1 + \frac{2^n+3^n}{4^n} \right)^{n\frac{2^n+3^n}{4^n}\frac{4^n}{2^n+3^n}} $$ Since $n\frac{2^n+3^n}{4^n}\to 0$ and $(1 + \frac{2^n+3^n}{4^n})^{\frac{4^n}{2^n+3^n}}\to e$ you have that the above limit is equal to $4e^0=4$.

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Thanks a lot sir.I have got it... –  learner Jan 8 '13 at 13:53
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