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

Question

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

Answer 1

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?

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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.