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
 A: 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.
