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

Note by Chinmay Sangawadekar 1 year, 1 month 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 · 1 year, 1 month ago

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@Deeparaj Bhat – Exactly !!! :) – Chinmay Sangawadekar · 1 year, 1 month ago

@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 · 1 year, 1 month ago

@Rishabh Tiwari – Try Elementary NT by David Burton – Chinmay Sangawadekar · 1 year, 1 month ago

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

@Deeparaj Bhat – The same way I did , its crct! – Rishabh Tiwari · 1 year, 1 month ago

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

@Ayush Rai – Yup . more standard solution is provided by Deepraj :) – Chinmay Sangawadekar · 1 year, 1 month ago

@Chinmay Sangawadekar – yup, he is far better than me!! – Ayush Rai · 1 year, 1 month ago

@Nihar Mahajan @Mehul Arora – Chinmay Sangawadekar · 1 year, 1 month ago

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## Comments

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

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– Chinmay Sangawadekar · 1 year, 1 month ago

Exactly !!! :)Log in to reply

@Chinmay Sangawadekar – Rishabh Tiwari · 1 year, 1 month ago

Can you please suggest me some Number Theory Books, I m a beginner & want to learn from basics to advanced , !Log in to reply

– Chinmay Sangawadekar · 1 year, 1 month ago

Try Elementary NT by David BurtonLog in to reply

– Rishabh Tiwari · 1 year, 1 month ago

Ok thank you very much , is it good for the very basics to advanced level & for olympiads?Log in to reply

– Rishabh Tiwari · 1 year, 1 month ago

The same way I did , its crct!Log in to reply

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

Since \(\frac{1}{\infty}=0; n^0=\boxed 1.\) – Ayush Rai · 1 year, 1 month ago

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– Chinmay Sangawadekar · 1 year, 1 month ago

Yup . more standard solution is provided by Deepraj :)Log in to reply

– Ayush Rai · 1 year, 1 month ago

yup, he is far better than me!!Log in to reply

@Nihar Mahajan @Mehul Arora – Chinmay Sangawadekar · 1 year, 1 month ago

Log in to reply