TITLE: Limits of a product
QUESTION [7 upvotes]: $N$ 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?

REPLY [4 votes]: Considering $$P_n=\prod_{i=1}^{n-1} \left(1+\frac i {n^2} \right)\implies \log(P_n)=\sum_{i=1}^{n-1} \log\left(1+\frac i {n^2} \right)$$ Since $n$ is large, by Taylor (or even using simple equivalents)
$$\log\left(1+\frac i {n^2} \right)=\frac{i}{n^2}-\frac{i^2}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ making 
$$\log(P_n)=\frac{(n-1) \left(6 n^2-2 n+1\right)}{12 n^3}+\cdots=\frac{1}{2}-\frac{2}{3 n}+O\left(\frac{1}{n^2}\right)$$ showing a limit of $\frac 12$. Then $P_n \to \sqrt e$.
Probably too early, but added for your curiosity, using Pochhammer symbols
$$P_n=\frac{\left(n^2+1\right){}_{n-1} } {n^{2n-2}}$$ Using asymptotics
$$\log(P_n)=\frac{1}{2}-\frac{2}{3 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor
$$P_n=e^{\log(P_n)}=\sqrt e\left(1-\frac{2}{3 n}+\frac{5}{9 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
It also allows a shortcut evaluation of the partial products even for small values of $n$ as shown below
$$\left(
\begin{array}{cccc}
 n & P_n & P_n \approx & \text{approximation} \\
 3 & \frac{110}{81} & 1.35802 & 1.38411 \\
 4 & \frac{2907}{2048} & 1.41943 & 1.43118 \\
 5 & \frac{570024}{390625} & 1.45926 & 1.46553 \\
 6 & \frac{1873495}{1259712} & 1.48724 & 1.49097 \\
 7 & \frac{20872566000}{13841287201} & 1.50799 & 1.51039 \\
 8 & \frac{418915572075}{274877906944} & 1.52401 & 1.52564 \\
 9 & \frac{316401132826240}{205891132094649} & 1.53674 & 1.53790
\end{array}
\right)$$