Let a be a positive number. Then $\lim_{n \to \infty}[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}]$ 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.
 A: We have
$$\sum_{k=1}^n \dfrac1{ka+n} = \dfrac1n \sum_{k=1}^n \dfrac1{1+a\cdot \dfrac{k}n} \sim \int_0^1 \dfrac{dx}{1+ax} = \dfrac{\log(1+a)}a$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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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\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}
