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# Is this correct?

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

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} · 6 months, 1 week ago

Exactly !!! :) · 6 months, 1 week ago

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

Try Elementary NT by David Burton · 6 months, 1 week ago

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

The same way I did , its crct! · 6 months, 1 week ago

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

Yup . more standard solution is provided by Deepraj :) · 6 months, 1 week ago

yup, he is far better than me!! · 6 months, 1 week ago