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Can we prove that for all natural numbers $n$, there exists a function $\epsilon _{n}$ satisyfing : $\forall t > -1, \sqrt[n]{1+t}=1+\frac{t.\epsilon_{n}(t)}{n}$ and $\lim_{t \to 0}\epsilon _{n}(t)=0$ ?

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up vote 1 down vote accepted

From the given equation, $$ \epsilon_n = \frac{n}{t}((1+t)^{\frac{1}{n}}-1) \to \frac{(1+t)^{\frac{1}{n}-1}}{1} \to 1$$

as $t \to 0$ by L'Hopital for $"\frac{0}{0}"$.

So the answer is No.

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But, by the binomial theorem, for fixed n, $(1+t)^{1/n} = 1 + t/n + (t^2/2)(1/n)(1/n-1) + ... = 1 + t/n + O(t^2)$ so that $\epsilon_n = 1 - t\frac{n-1}{n} + O(t^2) = 1 + O(t)$, not $0$.

Your computation $$\epsilon_n = \frac{n}{t}((1+t)^{\frac{1}{n}}-1) \to \frac{(1+t)^{\frac{1}{n}-1}}{1}$$ is correct, but the right-hand limit is $1$ as $t \to 0$, not $0$.

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Indeed, fixed it. –  Peteris Dec 12 '11 at 21:17
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