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?
 A: $$\left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right)=1+\frac{1}{n}+\frac{j(n-j)}{n^4}$$
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
$$\lim_n P[n]=\sqrt{e}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\prod_{k = 1}^{n - 1}\pars{1 + {k \over n^{2}}} & =
{1 \over n^{2n - 2}}\prod_{k = 1}^{n - 1}\pars{k + n^{2}} =
{1 \over n^{2n - 2}}\pars{1 + n^{2}}^{\overline{n - 1}} =
{1 \over n^{2n - 2}}
\,{\Gamma\pars{1 + n^{2} + n - 1} \over \Gamma\pars{1 + n^{2}}}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1 \over n^{2n - 2}}
\,{\root{2\pi}\pars{n + n^{2}}^{1/2 + n + n^{2}}\expo{-n - n^{2}} \over
\root{2\pi}\pars{1 + n^{2}}^{3/2 + n^{2}}\expo{-1 - n^{2}}}
\\[5mm] & =
{1 \over n^{2n - 2}}
\,{n^{1 + 2n + 2n^{2}}\pars{1 + 1/n}^{1/2 + n + n^{2}} \over
n^{3 + 2n^{2}}\pars{1 + 1/n^{2}}^{3/2+ n^{2}}}\,\expo{-n + 1}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\,{\pars{1 + 1/n}^{n} \over
\pars{1 + 1/n^{2}}^{n^{2}}}\,\bracks{\pars{1 + {1 \over n}}^{n^{2}}\expo{-n + 1}}
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\Large \to}\,\,\,
\lim_{n \to \infty}\exp\pars{n^{2}\ln\pars{1 + {1 \over n}} - n + 1} =
\expo{\color{red}{1/2}} = \bbx{\root{\expo{}}} \approx 1.6487
\end{align}

Note that
  $\ds{n^{2}\ln\pars{1 + {1 \over n}} - n + 1
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{=}\,\,\,
\color{red}{1 \over 2} + {1 \over 3n} + \mrm{O}\pars{1 \over n^{2}}}$.

A: Let $x$ be a real number in the interval $[0,1/2)$. Then we have the following known inequality:
$$ 1+x \le \exp x\ .$$
(Which holds for any real $x$.) This can be used to give an upper bound for the product, while also giving the chance to convert in the formula for the bound to a sum, which can be easily computed. We want a similar argument for a lower bound. For this, note that
$$
\frac 1{1+x}\le 1-x+x^2\le \exp(-x+x^2) = \frac 1{\exp(x-x^2)}\ .
$$
(We have applied the same inequality above for an other argument.) From the double inequality:
$$
\exp(x-x^2)\le 1+x\le \exp(x)
$$ we get
\begin{align}
\exp\sum_{0<k<n}\left(\frac k{n^2}-\frac {k^2}{n^4}\right)
&=
\prod_{0<k<n}\exp\left(\frac k{n^2}-\frac {k^2}{n^4}\right)
\\\\
&\le
\prod_{0<k<n}\left( 1+\frac k{n^2}\right)
\\\\
&\le
\prod_{0<k<n}\exp\frac k{n^2}
\\\\
&=
\exp\sum_{0<k<n}\frac k{n^2}
\ .
\end{align}
The expressions at the beginning and at the end enclose the given sequence and converge each to $\exp\frac 12$, which is thus also the limit of the given sequence.
Edit: Computer check using sage:
sage: R = RealField( 400 )
sage: N = 1000
sage: print "P[%s] ~ %.11f" % ( N, prod( [1+k/N^2 for k in [1..N-1] ] ) ) 
P[1000] ~ 1.64762303820
sage: print "exp(1/2) ~ %.11f" % R(exp(1/2))
exp(1/2) ~ 1.64872127070

A: 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.
