An ISI 2010 Problem

Let $a_1>a_2>.......>a_r$ be positive real numbers. Compute $\lim_{n \to \infty} ({a_1}^n +{a_2}^n + ........+ {a_r}^n)^{\frac{1}{n}}$ .

[Please be kind enough to write the solution.]

#NumberTheory #AlgebraicOperations

Easy Math Editor

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Comments

Let $L = \lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_r^x)^{1/x}.$ Then $\begin{aligned} \log L &= \log \lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_r^x)^{1/x} \\ &= \lim_{x \to \infty} \log (a_1^x + a_2^x + \dots + a_r^x)^{1/x} \\ &= \lim_{x \to \infty} \frac{\log (a_1^x + a_2^x + \dots + a_r^x)}{x}. \end{aligned}$

By L'Hopital's Rule, $\begin{aligned} \lim_{x \to \infty} \frac{\log (a_1^x + a_2^x + \dots + a_r^x)}{x} &= \lim_{x \to \infty} \frac{\frac{d}{dx} \log (a_1^x + a_2^x + \dots + a_r^x)}{\frac{d}{dx} x} \\ &= \lim_{x \to \infty} \frac{a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r}{a_1^x + a_2^x + \dots + a_r^x}. \end{aligned}$

Then $\begin{aligned} \lim_{x \to \infty} \frac{a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r}{a_1^x + a_2^x + \dots + a_r^x} &= \lim_{x \to \infty} \frac{(a_1^x \log a_1 + a_2^x \log a_2 + \dots + a_r^x \log a_r)/a_1^x}{(a_1^x + a_2^x + \dots + a_r^x)/a_1^x} \\ &= \lim_{x \to \infty} \frac{\log a_1 + (a_2/a_1)^x \log a_2 + \dots + (a_r/a_1)^x \log a_r}{1 + (a_2/a_1)^x + \dots + (a_r/a_1)^x} \\ &= \log a_1. \end{aligned}$

Therefore, $L = a_1$ .

Jon Haussmann - 7 years, 1 month ago

Thanks :)

Maharnab Mitra - 7 years, 1 month ago

Take a1 to power n common from bracket..it become lim(a1)[1 + (a2/a1)^n +....]^1/n .

It all terms terms become zero ..so answer will come to ...a1.

Aman Chaudhary - 7 years, 1 month ago

Not immediately true. The limiting value of $0 ^ 0$ depends on what path you take.

Edit: Ooops, I spoke too quickly. The power needs to be logarithmic. Use $f(x) = x, g(x) = \frac{ 1} {\log x}$ , then we have $f(x) ^ { g(x) } = e$ .

Calvin Lin Staff - 7 years, 1 month ago

$\lim_{x \to 0} x^{x^0.2} = e^{\lim_{x \to 0} x^{\frac{1}{5}} ln (x)}$

Now, $\lim_{x \to 0} x^{\frac{1}{5}} ln (x)= \lim_{x \to 0} \frac{ ln (x)}{x^{\frac{-1}{5}}}$ $=\lim_{x \to 0} \frac{-x^{\frac{1}{5}}}{5}$ (using $L'Hospital$ 's rule) which goes to 0. Thus, the required limit is $e^0=1$

Maharnab Mitra - 7 years, 1 month ago

@Maharnab Mitra – Sorry, updated.

Calvin Lin Staff - 7 years, 1 month ago

@Calvin Lin – Sir, then what should the correct answer (or rather solution) to the problem be?

Maharnab Mitra - 7 years, 1 month ago

Multiply and divide by a1 and proceed. Ans: a1

Pritiranjan Dwibedy - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$