Let a be a positive number. Then $\lim_{n \to \infty}[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}]$

Question

Problem :

Let $a$ be a positive number. Then $$\lim_{n \to \infty}\left[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}\right]$$

Please suggest how to proceed in such limit problems, will be of great help thanks.

Answer 1

We have $$\sum_{k=1}^n \dfrac1{ka+n} = \dfrac1n \sum_{k=1}^n \dfrac1{1+a\cdot \dfrac{k}n} \sim \int_0^1 \dfrac{dx}{1+ax} = \dfrac{\log(1+a)}a$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\pars{{1 \over a + n} + {1 \over 2a + n} + \cdots + {1 \over na + n}} = \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over ka + n} = {1 \over a}\,\lim_{n \to \infty}\sum_{k = 0}^{n - 1}{1 \over k + 1 + n/a} \\[5mm] = &\ {1 \over a}\,\lim_{n \to \infty}\sum_{k = 0}^{\infty} \pars{{1 \over k + 1 + n/a} - {1 \over k + n + 1 + n/a}} \\[5mm] = & {1 \over a}\,\lim_{n \to \infty}\pars{H_{n + n/a} - H_{n/a}}\qquad \pars{~H_{z}:\ Harmonic\ Number~} \end{align}

Since $\ds{H_{z} \sim \ln\pars{z} + \gamma + {1 \over 2z}\ \mbox{as}\ \verts{z} \to \infty\ \mbox{with}\ \,\verts{\mrm{arg}\pars{z}} < \pi}$ where $\ds{\gamma}$ is the Euler-Mascheroni Constant:

\begin{align} &\lim_{n \to \infty}\pars{{1 \over a + n} + {1 \over 2a + n} + \cdots + {1 \over na + n}} = {1 \over a}\,\lim_{n \to \infty}\ln\pars{n + n/a \over n/a} = \bbx{\ln\pars{1 + a} \over a} \end{align}