I have an upper Riemann Sum of a question which I have gotten to be $\sum_{i=1}^{n}\frac{n}{n^{2}+i^{2}}$ on the interval [0,1]. switching the $i$ to a summation I transformed this to $\lim_{n \to \infty} \frac{6}{2n^{2}+9n+1}.$ Using this result i need to show $\lim_{n \to \infty} \sum_{i=0}^{n-1} \frac{n}{n^{2}+j^{2}} = \frac{\pi}{4}$. I have no idea how i can show this any help? The original function is $F(x) = \frac{1}{1+x^{2}}$.
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$$\lim_{n\to +\infty}\sum_{j=0}^{n-1}\frac{n}{n^2+j^2} = \lim_{n\to +\infty}\frac{1}{n}\sum_{j=0}^{n-1}\frac{1}{1+\left(\frac{j}{n}\right)^2}=\int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}.$$
answered Jan 14, 2016 at 16:57
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$\begingroup$ what is j in this case? also would the limit of 0.5 be right in the first part? $\endgroup$ Jan 14, 2016 at 16:58
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$\begingroup$ @user305179: The summation index (I prefer to avoid $i$ since it is also the usual symbol for the imaginary unit). The second sum you wrote was $\sum_{i=0}^{n-1}\frac{n}{n^2+j^2}$ but $i$ does not appear in $\frac{n}{n^2+j^2}$. $\endgroup$ Jan 14, 2016 at 16:59
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$\begingroup$ @user305179: the limits $$\lim_{n\to +\infty}\sum_{j=0}^{n-1}\frac{n}{n^2+j^2}\quad\text{and}\quad \lim_{n\to +\infty}\sum_{j=1}^{n}\frac{n}{n^2+j^2}$$ are obviously the same. $\endgroup$ Jan 14, 2016 at 17:03