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I have been trying to evaluate the following limit

$$ \lim_{m \rightarrow \infty} \prod_{k = 0}^{m} \frac{n^{2^{k}} - 1}{n^{2^{k}}} $$

for fixed natural number $n > 1.$

Plugging some finite products into WolframAlpha it seems that the product decays very slowly to zero, but I'm unsure whether this limit actually converges to zero or to some other number. Any tips would be appreciated.

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  • $\begingroup$ You can simplify the fraction to get $1-1/n^{2^k}$. What did you try? As the product will always be $>0$, I would suggest finding an upper bound where you already know that it converges to $0$. $\endgroup$
    – LegNaiB
    May 14, 2021 at 22:15
  • $\begingroup$ I tried the $\epsilon-N$ definition by noticing that for all $m > N,$ $$|\prod_{k=0}^{m} \frac{n^{2^{k}} - 1}{n^{2^{k}}} - 0 | = |\prod_{k=0}^{m} \frac{n^{2^{k}} - 1}{n^{2^{k}}}| \leq |\prod_{k=0}^{N} \frac{n^{2^{k}} - 1}{n^{2^{k}}}|$$ and trying to bound this via epsilon. In terms of upper bounds, I am not sure what would work (the fact that it is a product is throwing me off). $\endgroup$
    – ksankar
    May 14, 2021 at 22:23
  • $\begingroup$ Would you like to find value of limit or only evaluating convergence is enough? $\endgroup$
    – zkutch
    May 14, 2021 at 22:35
  • $\begingroup$ I would like to find the value of the limit, or at the minimum show that it either is or isn't zero. $\endgroup$
    – ksankar
    May 14, 2021 at 22:37
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    $\begingroup$ Hint: a product of the form $\prod_k (1-t_k)$ (where $0<t_k<1$) converges to a nonzero number if and only if the series $\sum_k t_k$ converges. Indeed, it is for precisely this reason that we say a product like $\prod_{k=2}^\infty (1-\frac1k)$ diverges to $0$ rather than calling it a convergent product. $\endgroup$
    – Greg Martin
    May 14, 2021 at 23:16

1 Answer 1

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By definition infinite product $\prod\limits_{k=1}^{\infty}a_k$ converges to a nonzero real number if and only if the sum $\sum\limits_{k=1}^{\infty}\log a_k$ converges. In our case we can consider

$$\ln\left(1-\frac{1}{n^{2^k}}\right)=-\frac{1}{n^{2^k}}+\frac{1}{2n^{2^{k+1}}}+o\left(\frac{1}{n^{2^{k+1}}} \right)$$

where all series in right side converges.

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