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Calculate: $$ \lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2}+\frac{1}{x^2+2^2}+\dots+\frac{1}{x^2+x^2}\right)$$

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    $\begingroup$ Transform it into a Riemann sum. $\endgroup$ Jan 15, 2014 at 19:50
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    $\begingroup$ What have you tried yourself? Do you have any ideas about basic techniques for calculating limits and whether they work or not on this specific case? $\endgroup$
    – Arthur
    Jan 15, 2014 at 19:50
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    $\begingroup$ How? I have tried many times to solve this question. $\endgroup$
    – mathlover
    Jan 15, 2014 at 19:51
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    $\begingroup$ $x$ is a natural number? $\endgroup$ Jan 15, 2014 at 19:54
  • $\begingroup$ Yes, $ x $ is a natural number. $\endgroup$
    – mathlover
    Jan 15, 2014 at 19:56

2 Answers 2

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Idea: change the limit in a Riemann sum.

$$\lim_{n \to \infty}n \Bigg( \sum_{k=1}^n {\frac {1}{n^2+k^2}}\Bigg) = \lim_{n \to \infty}\frac{1}{n}\Bigg(\sum_{k=1}^n {\frac {n^2}{n^2+k^2}}\Bigg)= \lim_{n \to \infty}\frac{1}{n}\Bigg(\sum_{k=1}^n {\frac {1}{1+(\frac{n}{k})^2}}\Bigg)$$

$$=\int_{0}^1 {\frac {dx}{1+x^2}}=\arctan(1)-\arctan(0)=\frac{\pi}{4}$$

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$$ \begin{align} & \phantom{=} \lim_{x\to+\infty} x\left( \frac{1}{x^2+1^2} + \frac{1}{x^2+2^2} + \dots + \frac{1}{x^2+x^2}\right) \\[8pt] & = \lim_{x\to\infty} \frac 1 x \left( \frac{1}{1+\frac{1^2}{x^2}} + \frac{1}{1+\frac{2^2}{x^2}} +\cdots+\frac1{1+\frac{x^2}{x^2}} \right) \\[8pt] & = \int_0^1 \frac{1}{1+w^2} \, dw = \frac \pi 4. \end{align} $$

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