Riemann sum of log function

Question

Find the limit as $n \rightarrow \infty$ of \begin{align} {1 \over n}\,\log\left(\,{\left(\,2n\,\right)! \over n^{n}\,n!}\,\right) \end{align}

It's not in the form of a sum, so I am really confused as to what to do. Can someone please help?

Answer 1

$$\frac{1}{n}\log\left( \frac{(2n)!}{n^nn!}\right) = \frac{1}{n}\left[\log((2n)!) - \log(n!) - \log(n^n)\right] = $$ $$= \frac{1}{n}\left[\sum_{i=1}^{2n}\log(i) - \sum_{i=1}^n\log(i) - n\log(n)\right] =$$ $$= \frac{1}{n}\left[\sum_{i=1}^{n}\log(i) + \sum_{i=1}^{n}\log(n+i) - \sum_{i=1}^n\log(i) - \sum_{i=1}^n\log(n)\right] =$$ $$= \frac{1}{n}\sum_{i=1}^{n}\left[\log(i) + \log(n+i) - \log(i) - \log(n)\right] =$$ $$= \frac{1}{n}\sum_{i=1}^{n}\left[\log\left(\frac{n+i}{n}\right)\right].$$

Answer 2

We have that $$\begin{eqnarray*} \log\Bigl( \frac{(2n)!}{n^nn!}\Bigr) &=& \log( (2n)! ) - n \log(n) - \log n! \\&=& \sum_{k=1}^{2n} \log(k) - n \log n - \sum_{k=1}^{n} \log(k) \\&=& \sum_{k=n+1}^{2n} \log(k) - n \log(n) \\&=& \sum_{k=1}^{n} \Big( \log(n+k) - \log(n) \Big) \\&=& \sum_{k=1}^{n} \log( 1+ k/n) \end{eqnarray*}$$

So then

$$\begin{eqnarray*} \lim_{n\to\infty} \frac{1}{n} \log\Bigl( \frac{(2n)!}{n^nn!} \Bigr) &=&\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \log( 1+ k/n) \\&=& \int_0^1 \log(1+x) dx \\&=& \log(4) -1 \end{eqnarray*}$$

Answer 3

Using Stirling's approximation:

$$n! \sim \sqrt{2\pi n}\left ( \frac{n}{e} \right )^n$$

Since the limit is with $n \to \infty$ it is very useful in this case.

$$\begin{align} \lim_{n \to \infty} \frac 1n \ln \left ( \frac{(2n)!}{n^n n!} \right ) &= \lim_{n \to \infty} \frac 1n \ln \left ( \frac{\sqrt{4\pi n} \left ( \frac{2n}{e} \right )^{2n}}{n^n \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n} \right ) =\\ &= \lim_{n \to \infty} \frac 1n \ln \left ( \frac{\sqrt{2}\cdot(2n)^{2n}\cdot e^n}{n^n \cdot e^{2n} \cdot n^n} \right ) =\\ &= \lim_{n \to \infty} \frac 1n \ln \left ( \sqrt{2} \cdot \left ( \frac{4}{e} \right )^n \right ) =\\ &= \lim_{n \to \infty} \ln \left ( \sqrt{2} \cdot \left ( \frac{4}{e} \right )^n \right )^{\frac 1n} =\\ &= \lim_{n \to \infty} \ln 2^{\frac 1{2n}} + \lim_{n \to \infty} \ln4 + \lim_{n \to \infty} \ln e^{-1} =\\ &= \ln 4 - 1 \end{align}$$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#66f}{\large% \lim_{n\ \to\ \infty}\bracks{{1 \over n}\,\ln\pars{\pars{2n}! \over n^{n}\,n!}}} =\lim_{n\ \to\ \infty} {\bracks{2n\ln\pars{2n} - 2n} - n\ln\pars{n} - \bracks{n\ln\pars{n} - n} \over n} \\[3mm]&=\lim_{n\ \to\ \infty} \bracks{2\ln\pars{2} + 2\ln\pars{n} - 2 - \ln\pars{n} - \ln\pars{n} + 1} =\color{#66f}{\large 2\ln\pars{2} - 1} \approx {\tt 0.3863} \end{align}