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Is this correct?

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

Note by Chinmay Sangawadekar
6 months, 1 week ago

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\[\begin{align} \sqrt[\infty]{n}&=\lim_{k \to \infty} n^{\frac{1}{k}}\\ &=n^{\lim_{k \to \infty} \frac{1}{k} } \\ &=n^0\\&=1 \quad \forall n \in \mathbb{N} \end{align}\] Deeparaj Bhat · 6 months, 1 week ago

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@Deeparaj Bhat Exactly !!! :) Chinmay Sangawadekar · 6 months, 1 week ago

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@Chinmay Sangawadekar Can you please suggest me some Number Theory Books, I m a beginner & want to learn from basics to advanced , ! @Chinmay Sangawadekar Rishabh Tiwari · 6 months, 1 week ago

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@Rishabh Tiwari Try Elementary NT by David Burton Chinmay Sangawadekar · 6 months, 1 week ago

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@Chinmay Sangawadekar Ok thank you very much , is it good for the very basics to advanced level & for olympiads? Rishabh Tiwari · 6 months, 1 week ago

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@Deeparaj Bhat The same way I did , its crct! Rishabh Tiwari · 6 months, 1 week ago

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I think it is true.\(n^{\frac{1}{\infty}}\)
Since \(\frac{1}{\infty}=0; n^0=\boxed 1.\) Ayush Rai · 6 months, 1 week ago

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@Ayush Rai Yup . more standard solution is provided by Deepraj :) Chinmay Sangawadekar · 6 months, 1 week ago

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@Chinmay Sangawadekar yup, he is far better than me!! Ayush Rai · 6 months, 1 week ago

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@Nihar Mahajan @Mehul Arora Chinmay Sangawadekar · 6 months, 1 week ago

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