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Evaluate $$\lim_{n\to \infty} n^{-n^2}\left( \prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right) \right) ^n$$

Since on substituting $n=\infty$ we get a indeterminate form of $1^{\infty}$. Hence we can write the same limit as

$$\exp\left({\lim_{n\to \infty} \left(\frac {\prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right)}{n^{n-1} }-n \right)} \right)$$

Which evaluates to $e^{3/2}$ . Is it correct? I would also like to know if there are any other methods for this problem

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    $\begingroup$ Start by taking the log of the limit to convert the product into a sum and deal with some of the exponents and what-not, which I think is what you attempted to do, but forgot to take the log when you applied $\exp$ to the limit. $\endgroup$ – Simply Beautiful Art Apr 19 '18 at 16:41
  • $\begingroup$ Mathematica shows it is true $$\begin{align}&=\lim_{n\to\infty}n^{-n^2} \left(\prod _{r=0}^{n-1} \left(n+\frac{1}{3^r}\right)\right){}^n\\ &=\lim_{n\to\infty}n^{-n^2}\left(n^n \left(-\frac{1}{n};\frac{1}{3}\right)_n\right){}^n\\ &=\lim_{n\to\infty}\left(\left(-\frac{1}{n};\frac{1}{3}\right)_n\right){}^n\\ &=e^{3/2}\\ \end{align} $$ $\endgroup$ – John Glenn Apr 19 '18 at 17:15
  • $\begingroup$ @JohnGlenn so does the proof in my answer... Mathematica is good for checking or getting intuition, but it's not an oracle and doesn't amount to a proof. $\endgroup$ – Clement C. Apr 19 '18 at 17:16
  • $\begingroup$ @ClementC. That's why I didn't post it as an answer :) $\endgroup$ – John Glenn Apr 19 '18 at 17:32
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In your attempt, you rewrote the expression in an incorrect way (you forgot the logarithm, when rewriting $x = e^{\log x}$). How did you get that second expression?


You have $$ \left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n = \exp\left( n \sum_{r=0}^{n-1} \log(n+3^{-r})\right) = \exp\left( n^2\log n + n\sum_{r=0}^{n-1} \log(1+\frac{3^{-r}}{n})\right) \tag{1} $$ so that $$ n^{-n^2} \left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n = \exp\left( n\sum_{r=0}^{n-1} \log(1+\frac{1}{n3^r})\right)\,. \tag{2} $$ Now, you have $\log(1+u) = u+O(u^2)$ when $u\to 0$, from which $$\begin{align} n^{-n^2} \left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n &= \exp\left( n\sum_{r=0}^{n-1} \left(\frac{1}{n3^r}+ O\left(\frac{1}{n^23^r}\right)\right)\right) \\ &= \exp\left( \sum_{r=0}^{n-1} \left(\frac{1}{3^r}+ O\left(\frac{1}{n3^r}\right)\right)\right) \\ &= \exp \left( \frac{3}{2}(1+o(1)\right) \\ &\xrightarrow[n\to\infty]{} \boxed{e^{3/2}} \end{align}$$ indeed.

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  • $\begingroup$ Sorry there happened to be typo in the question. You might now check the question again. Sorry again $\endgroup$ – Rohan Shinde Apr 19 '18 at 16:45
  • $\begingroup$ Oh yes, that makes a quite big difference. $\endgroup$ – Clement C. Apr 19 '18 at 16:47
  • $\begingroup$ @Manthanein See my update. $\endgroup$ – Clement C. Apr 19 '18 at 16:58
  • $\begingroup$ Yep saw that. And yeah for my second expression I read a very special result for the indeterminate form of $1^{\infty}$. Let $f(x)$ and $g(x)$ be two functions such that $\lim_{x\to a} f(x)\to 1$ and $\lim_{x\to a} g(x)\to \infty$ then $$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} g(x)\cdot (f(x)-1)}$$ $\endgroup$ – Rohan Shinde Apr 19 '18 at 17:06
  • $\begingroup$ @Manthanein Ah, OK. I try not to use these blackbox statements in general, but yes, that's a valid method. $\endgroup$ – Clement C. Apr 19 '18 at 17:08

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