Solve \(\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}\)
Answer:
[(1/2)] tan-14 – 0
Solution:
\(\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=0}^{2n-1}\frac{1}{1+\frac{4r^2}{n^2}}=\int_0^2 \frac{1}{1+(2x)^2}dx\)
= [(1/2)] tan-14 – 0
