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

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.
