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?

$\frac{1}{n+a}+\dots+\frac{1}{n+an} = (\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+na})-(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+a-1})$
$= \log(n+na)-\log(n+a-1)+o(1) = \log(\frac{n(a+1)}{n+a-1})+o(1)$, so the limit as $n \to \infty$ is $\log(a+1)$, since $\log$ is continuous.
