Does $\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ converge? I want to check whether 
$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ 
converges or not. (a is a positive constant number.)
If it converges, how to find the value it converges?
And if not, why?
 A: Notice that we have a Riemann sum
$$\sum_{k=1}^n \frac{1}{n + ak} = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + a(k/n)}\to \int_0^1 \frac{dx}{1 + ax} = \frac{1}{a} \log(1+a)$$
A: This is the limit of a Riemann sum:
$$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na})=\frac{1}{a}\lim_{n\rightarrow\infty}\frac{a}{n}\sum_{k=1}^n\frac{1}{1+\frac{ka}{n}}\\=\frac{1}{a}\int_0^a\frac{dx}{1+x}=\frac{\ln(a+1)}{a}.$$
A: Since
$$
\sum_{k=1}^n\frac1k=\gamma+\log(n)+O\!\left(\frac1n\right)
$$
we get
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+ka}
&=\lim_{n\to\infty}\frac1a\sum_{k=1}^n\frac1{\frac na+k}\\
&=\frac1a\lim_{n\to\infty}\sum_{k=\frac na+1}^{\frac na+n}\frac1k\\
&=\lim_{n\to\infty}\frac1a\left[\log\left(\frac{\frac na+n}{\frac na}\right)+O\!\left(\frac1n\right)\right]\\[9pt]
&=\frac1a\log(a+1)
\end{align}
$$
A: $\frac{1}{n+a}+\dots+\frac{1}{n+an} = (\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+na})-(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+a-1})$
$= \log(n+na)-\log(n+a-1)+o(1) = \log(\frac{n(a+1)}{n+a-1})+o(1)$, so the limit as $n \to \infty$ is $\log(a+1)$, since $\log$ is continuous.
A: Subtracting out $1$ as $1/n$ from each term leaves $\sum_{k=1}^n\frac{-ka}{n(n+ka)}$. Cancelling the $-a$ factor gives a positive sum bounded above by $\sum_k\frac{k}{n^2}=\frac{1}{2}+o(1)$, so yes, the original limit converges.
