$\lim_{n \to \infty}(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})=e^{\frac{1}{2}}$. Here is the beginning of a proof:
Suppose $0<k \leq n$,
$1+\frac{1}{n}<(1+\frac{k}{n^2})(1+\frac{n+1-k}{n^2})=1+\frac{n+1}{n^2}+\frac{k(n+1-k)}{n^4}\leq 1+\frac{1}{n}+\frac{1}{n^2}+\frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above. 
 A: As an alternative, we have that
$$\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\ldots\left(1+\frac{n}{n^2}\right)=\prod_{k=1}^{n}\left(1+\frac{k}{n^2}\right)=e^{\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right) }$$
and
$$\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right)=\sum_{k=1}^{n}\left(\frac{k}{n^2}+k^2O\left(\frac{1}{n^4}\right)\right)=\frac1{n^2}\sum_{k=1}^{n}k+O\left(\frac{1}{n^4}\right)\sum_{k=1}^{n}k^2\to \frac12$$
indeed


*

*$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\implies \frac1{n^2}\sum_{k=1}^{n}k=\frac{n(n+1)}{2n^2}\to \frac12$

*$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\implies O\left(\frac{1}{n^4}\right)\sum_{k=1}^{n}k^2=O\left(\frac{1}{n}\right)\to 0$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\lim_{n \to \infty}\pars{1 + {1 \over n^{2}}}
\pars{1 + {2 \over n^{2}}}\cdots\pars{1 + {n \over n^{2}}} = \expo{1/2}:\ {\LARGE ?}}$.

\begin{align}
&\bbox[#ffd,10px]{\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}\prod_{k = 1}^{n}\pars{1 + {k \over n^{2}}}
\\[5mm] = &\
\lim_{n \to \infty}{\prod_{k = 1}^{n}\pars{k + n^{2}} \over
\prod_{k = 1}^{n}n^{2}} =
\lim_{n \to \infty}{\pars{1 + n^{2}}^{\large\overline{n}} \over \pars{n^{2}}^{n}}
\\[5mm] = &\
\lim_{n \to \infty}{\Gamma\pars{1 + n^{2} + n}/\Gamma\pars{1 + n^{2}} \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{-\pars{n^{2} + n}} \over
n^{2n}\bracks{\root{2\pi}\pars{n^{2}}^{n^{2} + 1/2}\expo{-n^{2}}}} \\[5mm] = &\
\lim_{n \to \infty}
{\pars{n^{2}}^{n^{2} + n + 1/2}\pars{1 + 1/n}^{n^{2} + n + 1/2}\,
\expo{-n} \over
n^{2n}\pars{n^{2n^{2} + 1}}}
\\[5mm] = &\
\lim_{n \to \infty}\exp\pars{\bracks{n^{2} + n + {1 \over 2}}
\ln\pars{1 + {1 \over n}} - n}
\\[5mm] = &\
\lim_{n \to \infty}\exp\pars{\bracks{n^{2} + n + {1 \over 2}}
\bracks{{1 \over n} - {1 \over 2n^{2}}} - n} = \bbx{\expo{1/2}}
\approx 1.6487
\end{align}

Note that
  
$\ds{\pars{n^{2} + n + {1 \over 2}}
\ln\pars{1 + {1 \over n}} - n
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1 \over 2} + {1 \over 3n} - {1 \over 6n^{2}}}$.

A: Hint: By AM-GM Inequality
$$\frac{k+ n+1-k}{2} \geq \sqrt{k(n+1-k)}$$
