Can someone please show me how to do such questions with necessary steps..$$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{6n}{9n^2-k^2}$$
I have seen a similar question, (Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ ) but couldn't understand the method used for solving. Thank you
$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{6n}{9n^2-k^2} = \frac{1}{n}\sum_{k=1}^{n}\frac{6n}{9n^2-k^2}= \frac{1}{n}\sum_{k=1}^{n}\frac{6}{9-(\frac{k}{n})^2}= \frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n}) = \int_0^1f(x)dx \, where\, f(x)= \frac{6}{9-x^2}$