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So I'm trying to solve the following limit:

$$\lim_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^3}\right)\dots \left(1-\frac{1}{n^n}\right)$$

Now, I tried getting the squeeze theorem around this one, since it does feel like something for the squeeze theorem. The upper bound is obviously $1$, but since each term decreases the product, it may seem like this approaches zero?

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    $\begingroup$ Where did you come by this expression? It converges to something in the neighborhood of $0.719$, since the multiplicands approach $1$ too quickly for the product to go to $0$. $\endgroup$
    – Brian Tung
    Dec 14, 2016 at 1:02
  • $\begingroup$ @BrianTung It is (apparently) from a calculus introductory course workbook. A friend asked me to solve this since he was not able to do it, and neither was I. $\endgroup$ Dec 14, 2016 at 1:03
  • $\begingroup$ Have you tried the standard trick of taking the $\log$ of it and analyzing it as a series? $\endgroup$ Dec 14, 2016 at 1:37
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    $\begingroup$ It isn't hard to prove convergence but I think it's unlikely a closed form exists. $\endgroup$
    – Sophie
    Dec 14, 2016 at 1:54
  • $\begingroup$ Is it possible the expression has been transcribed in error? $\endgroup$
    – Brian Tung
    Dec 14, 2016 at 2:28

1 Answer 1

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$$ \begin{align} \prod_{n=2}^\infty\left(1-\frac1{n^n}\right) &\ge\prod_{n=2}^\infty\left(1-\frac1{n^2}\right)\\ &=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}n\frac{n+1}n\\ &=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}n\prod_{n=2}^m\frac{n+1}n\\ &=\lim_{m\to\infty}\frac1m\frac{m+1}2\\[6pt] &=\frac12 \end{align} $$ So the product converges (that is, it is bounded away from $0$). However, numerically, the product is approximately $0.71915450096501024665446931$ and the Inverse Symbolic Calculator does not find anything for this number.

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  • $\begingroup$ Thank you. I asked for the original task from the textbook, and it had all 2nd degrees, not n-th, which is a completely different thing. However, I'll leave the question as it is, since this answer refers to it. $\endgroup$ Dec 15, 2016 at 10:03
  • $\begingroup$ @Transcendental: I have modified my answer. This one might be more useful to you. $\endgroup$
    – robjohn
    Dec 15, 2016 at 14:17
  • $\begingroup$ Oh, but I wanted to leave this question as it is, since it's a valid question in itself, and I believe it doesn't have an answer here. The edited version is already solved here, and it's a much less interesting problem. $\endgroup$ Dec 15, 2016 at 14:20
  • $\begingroup$ It answers the same question. However, it also answers your intended question. $\endgroup$
    – robjohn
    Dec 15, 2016 at 14:22
  • $\begingroup$ Oh, I just now noticed. Thanks. $\endgroup$ Dec 15, 2016 at 14:23

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