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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

Malu
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  • The terms within the summation, should be of the form, (1/n)*f(k/n) where k runs from 1 to n. After that, it can be evaluated in the form of an integral. Refer to conversion of limit of sum as an integral. – Srinath Prasad Apr 11 '20 at 15:45
  • @SrinathPrasad thank you – Malu Apr 11 '20 at 17:21

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$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{6n}{9n^2-k^2} = \frac{1}{n}\sum_{k=1}^{n}\frac{6n}{9n^2-k^2}= \frac{1}{n}\sum_{k=1}^{n}\frac{6}{9-(\frac{k}{n})^2}= \frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n}) = \int_0^1f(x)dx \, where\, f(x)= \frac{6}{9-x^2}$

Ravindra
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