<s>[INST] Limit of sequence and Riemann sum problem work verification I have to calculate $$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}}$$ My attempt: 
$$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}} = \lim_{n \to \infty}{\frac{1}{n^2}\frac{1}{n}\sum_{k=1}^{n}{\frac{1}{\frac{k^2}{n^2}-4}}} = \left(\lim_{n \to \infty}{\frac{1}{n^2}}\right) \cdot \int_0^1{\frac{1}{x^2-4}} = 0 \cdot \left(-\frac{\log(3)}{4}\right) =0$$
Am I right?
 [/INST] $$\sum_{k=1}^{n} \frac{n}{k^2-4n^2}=\sum_{k=1}^{n} \frac{n}{n^2\left(\frac{k^2}{n^2}-4\right)}=\frac{1}{n}\sum_{k=1}^{n} \frac{1}{\left(\frac{k^2}{n^2}-4\right)}$$
So 
$$\lim_{n \to\infty} \sum_{k=1}^{n} \frac{n}{k^2-4n^2}= \int_0^1 \frac{1}{x^2-4}\mathrm{d}x=\frac{-\ln(3)}{4} $$
I hope I make no mistake ! 
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