<s>[INST] 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?
 [/INST] First realize that, $$\prod_{i=1}^{n} 1+ \frac i{n^2} = \prod_{i=1}^n  \frac {n^2+i}{n^2} = \prod_{i=n^2+1}^{n^2+n}  \frac {i}{n^2}  $$
Then,
$$\prod_{i=n^2+1}^{n^2+n}  \frac {i}{n^2} = \exp \left (  \sum_{i=n^2+1}^{n^2+n} \log(i) - 2n\log(n) \right )$$
Then we use a Riemann inequality since $\log$ is an increasing function,
$$\int_{n^2}^{n^2+n} \log(x)dx  \leq \sum_{i=n^2+1}^{n^2+n} \log(i) \leq \int_{n^2+1}^{n^2+n+1} \log(x)dx  \mathrm{\ \ (Riemann \ inequality)}  $$ 
$$\begin{align}  \int_{n^2}^{n^2+n} \log(x)dx -2n\log(n)   & = [xlog(x)-x]_{n^2}^{n^2+n} -n\log(n^2) \\
& = (n^2+n) \log(n^2+n) -n^2 - n - n^2\log(n^2) +n^2 -n \log(n^2)\\
& = (n^2 +n) \log(1+\frac 1 n) - n  \\
& = (n^2 +n)(\frac 1 n - \frac 1 {2n^2} + O(n^{-3})) - n \underset{n\to \infty}\to \frac 12 \end{align} $$
Likewise
$$\begin{align}  \int_{n^2+1}^{n^2+n+1} \log(x)dx -2n\log(n)   & = [x\log(x)-x]_{n^2+1}^{n^2+n+1} -n\log(n^2) \\
& = (n^2+1)\log(1+\frac{n}{n^2+1})+n\log(1+\frac{n+1}{n^2})-n \underset{n\to \infty}\to \frac 12  \end{align} $$
Therefore $\log(P_n) \underset{n\to \infty}\to \frac 12$ and $P_n \underset{n\to \infty}\to \sqrt e$
The only result I used is $x\mathcal \in V(0), \ \log(1+x)=x-\frac 12 x^2+O(x^3)$ to take the limits. Thanks to @achille hui  for the correction.
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