Converting a complex Riemann sum into a definite integral

Question

I'm having some trouble converting this Riemann sum into a definite integral on the interval [0,1]. Any help/explanations would be much appreciated. Thank you.

$$\sum_{i=1}^n \frac{1}{n(2+\frac in)ln(2+\frac in)}$$

Answer 1

Hint. One may obtain, as $n \to \infty$, $$ \sum_{i=1}^n \frac{1}{n(2+\frac in)\ln(2+\frac in)} \to \int_0^1\frac{dx}{(2+x)\ln(2+x)}=\int_0^1\frac{(\ln(2+x))'}{\ln(2+x)}\:dx $$ where we have applied the Riemann sum result $$ \sum_{i=1}^n \frac1n f\left(\frac in\right) \to \int_0^1f(x)\:dx $$ to $$ f(x)=\frac{1}{(2+x)\ln(2+x)}. $$

Answer 2

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

If you like an 'intuitive approach', write: \begin{align} {i \over n} & \equiv x_{i}\quad\mbox{and}\quad \pars{~\Delta x \equiv x_{i + 1} - x_{i} = {1 \over n} \implies n\Delta x = 1~} \end{align}

Your sum becomes: \begin{align} &\sum_{i = 1}^{n}{1 \over n\pars{2 + i/n}\ln\pars{2 + i/n}} = \sum_{i = 1}^{n}{1 \over n\pars{2 + x_{i}}\ln\pars{2 + x_{i}}}\,n\Delta x = \sum_{x_{i}\ \in\ \Omega}{\Delta x \over \pars{2 + x_{i}}\ln\pars{2 + x_{i}}} \\ & \mbox{where}\quad\Omega \equiv \braces{{1 \over n},{2 \over n},\ldots,1} \end{align} As $\ds{n \to \infty}$, $\ds{\Delta x \to 0}$ such that you'll arrive to an integral. The rest of the history is given in $\texttt{@Olivier Oloa}\,\,\,$ answer.