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

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?

• 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 $0$. – Brian Tung Dec 14 '16 at 1:02
• @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. – Transcendental Dec 14 '16 at 1:03
• Have you tried the standard trick of taking the $\log$ of it and analyzing it as a series? – Fimpellizieri Dec 14 '16 at 1:37
• It isn't hard to prove convergence but I think it's unlikely a closed form exists. – Sophie Dec 14 '16 at 1:54
• Is it possible the expression has been transcribed in error? – Brian Tung Dec 14 '16 at 2:28

1 Answer

\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.

• 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. – Transcendental Dec 15 '16 at 10:03
• @Transcendental: I have modified my answer. This one might be more useful to you. – robjohn Dec 15 '16 at 14:17
• 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. – Transcendental Dec 15 '16 at 14:20
• It answers the same question. However, it also answers your intended question. – robjohn Dec 15 '16 at 14:22
• Oh, I just now noticed. Thanks. – Transcendental Dec 15 '16 at 14:23