6
$\begingroup$

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?

| cite | improve this question | |
$\endgroup$
  • 2
    $\begingroup$ 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 '16 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. $\endgroup$ – Transcendental Dec 14 '16 at 1:03
  • $\begingroup$ Have you tried the standard trick of taking the $\log$ of it and analyzing it as a series? $\endgroup$ – Fimpellizieri Dec 14 '16 at 1:37
  • 6
    $\begingroup$ It isn't hard to prove convergence but I think it's unlikely a closed form exists. $\endgroup$ – Sophie Dec 14 '16 at 1:54
  • $\begingroup$ Is it possible the expression has been transcribed in error? $\endgroup$ – Brian Tung Dec 14 '16 at 2:28
2
$\begingroup$

$$ \begin{align} \prod_{n=2}^\infty\left(1-\frac1{n^n}\right) &\ge\prod_{n=2}^\infty\left(1-\frac1{n^2}\right)\\ &=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}n\frac{n+1}n\\ &=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}n\prod_{n=2}^m\frac{n+1}n\\ &=\lim_{m\to\infty}\frac1m\frac{m+1}2\\[6pt] &=\frac12 \end{align} $$ 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.