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Let $(a_{n})$ be a sequence of positive reals. If $a_{n}\rightarrow l\neq 0$ then $(a_{1}a_{2}\dots a_{n})^{\frac{1}{n}}$ also converges to l.

I know how to prove this result by taking log, using Cauchy's first theorem on limits, continuity of log etc.

I tried proving it by just using definition of convergence and didn't succeed. Is it possible to prove the result by just using definition of convergence (i.e. $\forall\epsilon>0(\exists N\in\mathbb{N}$ such that $\forall n\geq N(x_{n}\in(l-\epsilon,l+\epsilon)))$)

Shash
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  • This has been asked and answered several times, see https://math.stackexchange.com/q/1220020/42969 and the linked questions. – Martin R Jul 01 '23 at 17:51
  • @MartinR Nowhere in these links can I find an answer to the OP's precise question about a direct proof (not using log)/ – Anne Bauval Jul 01 '23 at 18:15
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    @AnneBauval: What about this: https://math.stackexchange.com/a/4028061/42969 ? – Martin R Jul 01 '23 at 18:49

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