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\[\sqrt[\infty]{n} =1\] where \(n\) belongs to Natural numbers .

Note by Chinmay Sangawadekar 1 year, 8 months ago

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\[\begin{align} \sqrt[\infty]{n}&=\lim_{k \to \infty} n^{\frac{1}{k}}\\ &=n^{\lim_{k \to \infty} \frac{1}{k} } \\ &=n^0\\&=1 \quad \forall n \in \mathbb{N} \end{align}\]

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Exactly !!! :)

Can you please suggest me some Number Theory Books, I m a beginner & want to learn from basics to advanced , ! @Chinmay Sangawadekar

@Rishabh Tiwari – Try Elementary NT by David Burton

@Chinmay Sangawadekar – Ok thank you very much , is it good for the very basics to advanced level & for olympiads?

The same way I did , its crct!

I think it is true.\(n^{\frac{1}{\infty}}\) Since \(\frac{1}{\infty}=0; n^0=\boxed 1.\)

Yup . more standard solution is provided by Deepraj :)

yup, he is far better than me!!

@Nihar Mahajan @Mehul Arora

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`\sin \theta`

`\boxed{123}`

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TopNewest\[\begin{align} \sqrt[\infty]{n}&=\lim_{k \to \infty} n^{\frac{1}{k}}\\ &=n^{\lim_{k \to \infty} \frac{1}{k} } \\ &=n^0\\&=1 \quad \forall n \in \mathbb{N} \end{align}\]

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Exactly !!! :)

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Can you please suggest me some Number Theory Books, I m a beginner & want to learn from basics to advanced , ! @Chinmay Sangawadekar

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The same way I did , its crct!

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I think it is true.\(n^{\frac{1}{\infty}}\)

Since \(\frac{1}{\infty}=0; n^0=\boxed 1.\)

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Yup . more standard solution is provided by Deepraj :)

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yup, he is far better than me!!

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@Nihar Mahajan @Mehul Arora

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