Limits of a product $N$ is a positive integer.
I want to calculate this limit, but i couldn't get anywhere when i tried.
$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$
I tried to apply $\ln()$ at both sides to transform into a sum etc..
tried to use functions to limit the superior and inferior intervals of the function like this:
$$\exp\left(\left(n-1\right)\ln\left(1+\frac1{n^2}\right)\right)<P[n]<\exp\left((n-1)\ln\left(1+\frac{n-1}{n^2}\right)\right)$$
The only thing I get is that:
$1<\lim(P[n])<e$
Can anyone help me to solve this?

$$\left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right)=1+\frac{1}{n}+\frac{j(n-j)}{n^4}$$
By AM-GM we have 
$$\sqrt{j(n-j)}\leq \frac{n}{2} $$
and hence
$$1+\frac{1}{n} \leq \left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right) \leq 1+\frac{1}{n}+\frac{1}{4n^2}$$
Therefore
$$\left(1+\frac{1}{n}\right)^{\frac{n-1}{2}} \leq P[n] \leq \left(  1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2}$$
Now
$$\lim_n \left(1+\frac{1}{n}\right)^{\frac{n-1}{2}}=\left( \left(1+\frac{1}{n}\right)^{n} \right)^{\frac{n-1}{2n}}=\sqrt{e}$$
And
$$\lim_n \left(  1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2} =\lim_n \left(\left(  1+\frac{4n+1}{4n^2} \right)^\frac{4n^2}{4n+1}\right)^\frac{4n+1}{8n} =\sqrt{e}$$
Therefore
$$\lim_n P[n]=\sqrt{e}$$
