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

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

• 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. – Simply Beautiful Art Apr 19 '18 at 16:41
• 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} – John Glenn Apr 19 '18 at 17:15
• @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. – Clement C. Apr 19 '18 at 17:16
• @ClementC. That's why I didn't post it as an answer :) – John Glenn Apr 19 '18 at 17:32

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.
• 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 $$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} g(x)\cdot (f(x)-1)}$$ – Rohan Shinde Apr 19 '18 at 17:06