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One of my friends asked me this question while doing a proof for convergence and this has confused me quite a bit. While this can be proved when $\frac{1}{\alpha} \in \mathbb{Z}$ easily by algebra of limits (specifically, by the result: $\{a_n\}\rightarrow a$ and $\{b_n\}\rightarrow b \Rightarrow \{a_nb_n\} \rightarrow ab$), I am not sure how to go about proving this result when $\frac{1}{\alpha} \in (0,1)$

One approach that I could come up with was by using the fact that the function $ f(x) = x^{\frac{1}{\alpha}}\ $ is continuous everywhere in its domain $\mathbb{R}^{+}$ and hence for every sequence $\{x_n\} \rightarrow a \; , \{f(x_n)\}\rightarrow f(a),$ therefore if $\{x_n\}\rightarrow a$ then $ \{f(x_n)\} = \{x_n^{\frac{1}{\alpha}}\}\rightarrow f(a) = a^\frac{1}{\alpha} $.

However, I and my friend too, are not comfortable using the concept of continuity to show the result and are looking for purely "sequences" based proof.

Also, my intuition tells me that the Bernoulli inequality might come in handy but I can't really pin it down.

Thank you!

Yash Burman
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    Various proofs here: https://math.stackexchange.com/q/1394697/42969, and here: https://math.stackexchange.com/q/3445618/42969, and here: https://math.stackexchange.com/q/2796186/42969 – Martin R Oct 25 '24 at 01:58
  • Are you assuming that $\alpha$ is an integer? If so, please edit your post to specify it. If not, better choose a simpler notation than $1/\alpha$ for your exponent. – Anne Bauval Oct 25 '24 at 02:47
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    Don't use the continuity of $f$, because that's what you're trying to prove. If you're not just considering $\alpha \in \Bbb{N}$, or indeed $\alpha \in \Bbb{Q} \setminus {0}$, then you will need to consider what exponentiation actually means when the exponent is irrational, because it doesn't follow so obviously from what $x^{p/q}$ means where $p, q \in \Bbb{Z}$. The most common way to do this is to define $x^y := \exp(y \ln x)$, where $\exp$ is the natural exponential $e^x$ (defined either by Taylor series or $\exp(x):=\lim_{n\to\infty}(1+x/n)^n$), and $\ln$ is its inverse. – Theo Bendit Oct 25 '24 at 03:58
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    You can use $$0\leq\left|(x_n)^{\frac{1}{\alpha}} - a^{\frac{1}{\alpha}}\right|\leq|x_n - a|^{\frac{1}{\alpha}}.$$ – Riemann Oct 25 '24 at 09:13
  • @TheoBendit could you please guide me to some additional resources/topics/hints related to proving this result for $\alpha \in \mathbb{Q} \backslash {0}$. Moreover can you explain why/how does irrational exponents come into the picture when I am trying to prove this result for $\alpha \in \mathbb{Q} \backslash {0}$ ?
    Thank you!     – Yash Burman Oct 25 '24 at 13:00
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    @YashBurman If you want $\alpha$ to be rational, then my comment need not be relevant. A proper definition of exponentials, logarithms, and real exponents can be found in any introductory real analysis course. – Theo Bendit Oct 25 '24 at 15:01

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Use the inequality:$|x^{p}-y^{p}| \leq |x-y|^p$, we know that $$\left|(x_n)^{\frac{1}{\alpha}}-a^{\frac{1}{\alpha}}\right|\leq|x_n - a|^{\frac{1}{\alpha}}.$$ For any given $\epsilon>0$, there exists $N$ such that, when $n>N$, we have $$|x_n-a|<\epsilon^{\alpha}.$$ So $$\left|(x_n)^{\frac{1}{\alpha}}-a^{\frac{1}{\alpha}}\right|\leq|x_n - a|^{\frac{1}{\alpha}}<\epsilon.$$ Hence, $$\lim_{n\to\infty}(x_n)^{\frac{1}{\alpha}}=a^{\frac{1}{\alpha}}.$$

Riemann
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