How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$? How can we compute the following limit:

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$

Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\lim_{n \to \infty}\pars{1 + {1 \over n^{2}}}
\pars{1 + {2 \over n^{2}}}\cdots\pars{ 1 + {n \over n^{2}}} =
\lim_{n \to \infty}{\pars{n^{2} + 1}\pars{n^{2} + 2}\cdots\pars{n^{2} + n} \over n^{2n}}
\\[5mm] = &\
\lim_{n \to \infty}{\pars{n^{2} + 1}^{\large\overline{n}} \over n^{2n}} =
\lim_{n \to \infty}
{\Gamma\pars{n^{2} + 1 + n}/\Gamma\pars{n^{2} + 1} \over n^{2n}} =
\lim_{n \to \infty}
{\pars{n^{2} + n}! \over n^{2n}\,\pars{n^{2}}!}
\\[5mm] = &\
\lim_{n \to \infty}
{\root{2\pi}\pars{n^{2} + n}^{n^{2} + n + 1/2}\expo{-n^{2} - n} \over
n^{2n}\bracks{\root{2\pi}\pars{n^{2}}^{n^{2} + 1/2}\expo{-n^{2}}}} =
\lim_{n \to \infty}
{n^{2n^{2} + 2n + 1}\ \pars{1 + 1/n}^{n^{2} + n + 1/2}\ \expo{-n} \over
n^{2n}\pars{n^{2n^{2} + 1}}}
\\[5mm] = &\
\exp\pars{\lim_{n \to \infty}\braces{\bracks{n^{2} + n + {1 \over 2}}\ln\pars{1 + {1 \over n}} - n}}
\\[5mm] = &\
\exp\pars{\lim_{n \to \infty}\braces{\bracks{n^{2} + n + {1 \over 2}}
\bracks{{1 \over n} - {1 \over 2n^{2}}} - n}}
\\[5mm] = &\
\exp\pars{\lim_{n \to \infty}\braces{%
\bracks{n -{1 \over 2} + 1 - {1 \over 2n} + {1 \over 2n} - {1 \over 4n^{3}}} - n}} = \bbx{\ds{\root{\expo{}}}}\\ &
\end{align}
A: For $1\le k\le n$, we have
$$\begin{align}
\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}
&=\int_1^{1+k/n^2}{dt\over t}-{k\over n(n+1)}\\
&=\int_0^{k/n^2}\left({1\over1+t}-{n\over n+1}\right)dt\\
&={1\over n+1}\int_0^{k/n^2}\left(1-tn\over1+t \right)dt\\
&\le{1\over n+1}\int_0^{k/n^2}dt\\
&\le{k\over n^2(n+1)}\\
&\le{1\over n(n+1)}
\end{align}$$
Note, the integral in the third line is clearly non-negative, which means the initial expression is also non-negative.  Therefore
$$0\le\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\le{1\over n(n+1)}$$
Consequently, since $1+2+\cdots+n={n(n+1)\over2}$, we have
$$0\le\sum_{k=1}^n\log\left(1+{k\over n^2}\right)-{1\over 2}=\sum_{k=1}^n\left(\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\right)\le{1\over n+1}$$
so by the Squeeze Theorem
$$\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+{k\over n^2}\right)={1\over2}$$
and thus
$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)=e^{1/2}=\sqrt e$$
A: We will estimate
$$f(n) = \sum_{k=1}^n \log \left(1 + \dfrac{k}{n^2}\right)$$
Recall that $\log(1+x) = x + \mathcal{O}(x^2)$. Hence, we get that
$$f(n) = \sum_{k=1}^n \left(\dfrac{k}{n^2} + \mathcal{O}\left(\dfrac{k^2}{n^4}\right)\right) = \dfrac{n+1}{2n} + \mathcal{O}(1/n) \tag{$\star$}$$
where we made use of the fact that $\displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}2$ and $\displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}6$.
Now letting $n \to \infty$ in $(\star)$, we get that
$$\lim_{n \to \infty} f(n) = \dfrac12$$
The product you are interested in is $e^{f(n)}$ and since $e^x$ is continuous, we have
$$\lim_{n \to \infty} e^{f(n)} = e^{\lim_{n \to \infty} f(n)} = e^{1/2} = \sqrt{e}$$
A: Using
$$\lim_{n\to\infty}\left(1+\frac an\right)^n=e^a$$
We have
\begin{align}
L&=\lim_{n\to\infty} \left[\left((1+\frac{1}{n^2})(1+\frac{2}{n^2})(1+\frac{3}{n^2})\cdots(1+\frac {n}
{n^2})\right)^{n^2}\right]^{\frac1{n^2}}\\
&=\lim_{n\to\infty}e^{\frac{1+\ldots+n}{n^2}}=\lim_{n\to\infty}e^\frac{n(n+1)}{2n^2}=\sqrt e
\end{align}
