$\lim_{n \to \infty}(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})=e^{\frac{1}{2}}$.

Question

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.

Answer 1

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

Answer 2

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

• $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\implies \frac1{n^2}\sum_{k=1}^{n}k=\frac{n(n+1)}{2n^2}\to \frac12$
• $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\implies O\left(\frac{1}{n^4}\right)\sum_{k=1}^{n}k^2=O\left(\frac{1}{n}\right)\to 0$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}\pars{1 + {1 \over n^{2}}} \pars{1 + {2 \over n^{2}}}\cdots\pars{1 + {n \over n^{2}}} = \expo{1/2}:\ {\LARGE ?}}$.

\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

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